Question #72cfd

You can also look at Steve 's and Truong-Son N. 's responses for some different information; here is what I have:

We're asked to find the ratio of the horizontal ranges of two particles after the initial speed if particle A is doubled, knowing that before it is doubled, the ratio of maximum heights reached of A to B is #3:1#.

To solve this, we can first find an equation for the maximum height reached of a particle:

We can use the equation

#(v_y)^2 = (v_0sinalpha_0)^2 - 2gh#

and solve for the height, #h#, knowing that at its maximum height, the instantaneous velocity #v_y# is #0#:

#0 = (v_0sinalpha_0)^2 - 2gh#

#h = ((v_0sinalpha_0)^2)/(2g)#

Calling the initial launch angles of particles A and B #alpha_A# and #alpha_B#, and knowing that the ratio of #h_A# to #h_B# is #3:1#, we can set up the following relationship:

#(h_A)/(h_B) = (((v_0sinalpha_A)^2)/(2g))/(((v_0sinalpha_B)^2)/(2g)) = ((sinalpha_A)^2)/((sinalpha_B)^2) = 3/1 = 3#

#(sinalpha_A)/(sinalpha_B) = sqrt3#

Now, let's derive an equation for the horizontal range #R# of a particle, using the equations

#h = (v_0sinalpha_0)t - 1/2g t^2#

and

#R = (v_0cosalpha_0)t#

Since the height #h# is #0# when it lands again, we solve for the time, #t#:

#0 = (v_0sinalpha_0)t - 1/2g t^2#

#1/2g t = v_0sinalpha_0#

#t = (2v_0sinalpha_0)/g#

Now we can plug this value in for #t# in the second equation:

#R = (v_0cosalpha_0)((2v_0sinalpha_0)/g)#

#R = (2(v_0)^2sinalpha_0cosalpha_0)/g#

This equation for each particle A and B is

#R_A = (2(v_0)^2sinalpha_Acosalpha_A)/g#

#R_B = (2(v_0)^2sinalpha_Bcosalpha_B)/g#

We're trying to find the ratio #(R_A)/(R_B)# after the speed is doubled, so we can write it as

#(R_A)/(R_B) = ((2(color(blue)(2)v_0)^2sinalpha_Acosalpha_A)/g)/((2(v_0)^2sinalpha_Bcosalpha_B)/g) = ((color(blue)(2)v_0)^2sinalpha_Acosalpha_A)/((v_0)^2sinalpha_Bcosalpha_B)#

#= (color(blue)(4)sinalpha_Acosalpha_A)/(sinalpha_Bcosalpha_B)#

From the first part, we found that

#(sinalpha_A)/(sinalpha_B) = sqrt3#

so we now have

#(R_A)/(R_B) = color(red)((4sqrt3cosalpha_A)/(cosalpha_B)#

Since the cosine ratio can be an infinite number of quantities, this is as simplified as it gets.

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

#{: (v=u+at, " where ", s="displacement "(m)), (s=ut+1/2at^2, , u="initial speed "(ms^-1)), (s=1/2(u+v)t, , v="final speed "(ms^-1)), (v^2=u^2+2as, , a="acceleration "(ms^-2)), (s=vt-1/2at^2, , t="time "(s)) :} #

Initial Projection:

Let us start with what we know initially about the particles #A# and #B#. They are both projected with the same speed, let us call the initial speeds #u#.

We also know that #A# and #B# attain different heights with a ratio #3:1#, thus they must be projected at different angles.

Let us denote the height by #h_a#, #h_b# and angles by #theta_a#, #theta_b# respectively. Then:

# h_a = 3h_b #

Initial Projection - Vertical Motion

The projectiles travels under constant acceleration due to gravity. Its speed will be instantaneous #0# at the maximum of its trajectory. Considering upwards as positive, and resolving the initial speed vertically, we have:

For particle A:

# { (s=,h_a,m),(u=,u sin (theta_a),ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,"----",s) :} #

Applying #v^2=u^2+2as# we have:

# 0=u^2 sin^2 (theta_a) + 2(-g)(h_A) => h_a = (u^2 sin^2 (theta_a))/(2g) #

For particle B:

# { (s=,h_b,m),(u=,u sin (theta_b),ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,"----",s) :} #

Applying #v^2=u^2+2as# we have:

# 0=u^2 sin^2 (theta_b) + 2(-g)(h_b) => h_b = (u^2 sin^2 (theta_b))/(2g) #

Then using # h_a = 3h_b# we have:

# (u^2 sin^2 (theta_a))/(2g) = 3 (u^2 sin^2 (theta_b))/(2g) #

# :. sin^2 (theta_a) = 3 sin^2 (theta_b) #

Secondary Projection:

The speed of #A# is doubled, all other parameters remain the same. we want to calculate the ratio of the new horizontal ranges.

Secondary Projection - Vertical Motion

We now use the fact that to perform the full trajectory and hit the ground again, the total displacement in height is zero.

In order to calculate the horizontal displacements, say #x_a# and #x_b# we will need the time each particle is in motion. Let use denote these times by #T_a# and #T_b#

For particle A:

# { (s=,0,m),(u=,2u sin (theta_a),ms^-1),(v=,"-----",ms^-1),(a=,-g,ms^-2),(t=,T_a,s) :} #

Applying #s=ut+1/2at^2# we have:

# 0=2u sin (theta_a) T_a + 1/2(-g)T_a^2 #
# :. T_a(2u sin (theta_a) -1/2gT_a) = 0#
# :. T_a = 0, (4u sin (theta_a))/g #

For particle B:

# { (s=,0,m),(u=,u sin (theta_b),ms^-1),(v=,"-----",ms^-1),(a=,-g,ms^-2),(t=,T_b,s) :} #

Applying #s=ut+1/2at^2# we have:

# 0=u sin (theta_b) T_b + 1/2(-g)T_b^2 #
# :. T_b(u sin (theta_b) -1/2gT_b) = 0#
# :. T_b = 0, (2u sin (theta_b))/g #

We can ignore both solution #T_a=0# and #T_b=0# as this the time at the point the particles are released. Thus we have found our two times of motion #T_a# and #T_b# for #A# and #B# respectively.

Secondary Projection - Horizontal Motion

Having found the times of motion we must now calculate the horizontal distance traveled by each particle so that we can form their ratio as asked by the question.

The particles will move under constant speed (NB we can still use "suvat" equation with a=0). We must resolve the initial speed in the horizontal direction:

For particle A:

# { (s=,x_a,m),(u=,2u cos (theta_a),ms^-1),(v=,"-----",ms^-1),(a=,0,ms^-2),(t=,T_a,s) :} #

So applying #s=ut+1/2at^2 => s=ut \ # we get

# x_a = 2u cos (theta_a) (4u sin (theta_a))/g#
# \ \ \ = (8u^2 cos(theta_a)sin (theta_a))/g#

For particle B:

# { (s=,x_b,m),(u=,u cos (theta_b),ms^-1),(v=,"-----",ms^-1),(a=,0,ms^-2),(t=,T_b,s) :} #

So applying #s=ut+1/2at^2 => s=ut \ # we get

# x_b = u cos (theta_b) (2u sin (theta_b))/g #
# \ \ \ = (2u^2 cos(theta_b)sin (theta_b))/g #

Summing Up

We have established that:

# sin^2 (theta_a) = 3 sin^2 (theta_b) => sin^2 (theta_a)/sin^2 (theta_b) = 3#
# x_a = (8u^2 cos(theta_a)sin (theta_a))/g#
# x_b = (2u^2 cos(theta_b)sin (theta_b))/g #

We are asked in the question to find the ratio of the horizontal ranges, that is the ratio:

# x_a / x_b = ( (8u^2 cos(theta_a)sin (theta_a))/g ) / ((2u^2 cos(theta_b)sin (theta_b))/g) #

# " " = ( (8u^2 cos(theta_a)sin (theta_a))/g ) * (g/(2u^2 cos(theta_b)sin (theta_b))) #

# " " = 4 \ (cos(theta_a)sin (theta_a)) /(cos(theta_b)sin (theta_b)) #

# " " = 4 \ (2cos(theta_a)sin (theta_a)) /(2cos(theta_b)sin (theta_b)) = 4 \ (sin (2theta_a)) /(sin (2theta_b)) #

Although we know the ratio:

# sin^2 (theta_a)/sin^2 (theta_b) = 3 => sin (theta_a)/sin (theta_b) = +- sqrt(3) #
(NB See Additional Observations notes below)

we do not know the ratio of:

#cos (theta_a)/cos (theta_b)# or #sin (2theta_a)/sin (2theta_b)#

and there does not appear to be relationship between these entities, which prevents the solution from progressing.

Additional Observations

One further observation is that:

# 90^o lt theta_a lt 0^o # and # 90^o lt theta_b lt 0^o #

The case #theta_a=0^o# or #theta_b=0^o# is specifically excluded as this would have horizontal motion only, and vertical motion is required by the problem to allow the height ratio to exist.

Similarly, the case #theta_a=90^o# or #theta_b=90o# is also specifically excluded as this would have vertical motion only, and horizontal motion is required by the problem to allow the range ratio to exist.

With this in mind, we can show that from the derived condition:

# sin^2 (theta_a)/sin^2 (theta_b) = 3 #

Combined with the angle restrictions that we can ignore the negative solution, and so:

# sin (theta_a)/sin (theta_b) = sqrt(3) => sin (theta_a) = sqrt(3) sin (theta_b)#

Now, we require that:

# | sin (theta_a) | le 1 => | sqrt(3) sin (theta_b) | le 1#
# :. | sin (theta_b) | le 1/sqrt(3)#
# :. theta_b le 35.26^o #

Similarly:

# | sin (theta_b) | le 1 => | (sin (theta_a))/sqrt(3) | le 1#
# :. | sin (theta_a) | le sqrt(3)#

Which is always satisfied.

From which we conclude that there is a restriction on the launch angles #theta_a# and #theta_b# that we introduced such that:

# 90^o lt theta_a lt 0^o \ \ # and # \ \ 35.26^o lt theta_b lt 0^o #

Nathan and Steve have great answers. I thought I'd provide some interesting data though. As they've gotten, one of the simpler expressions for the ratio of the horizontal ranges is:

#R_A/R_B = (v_A/v_B)^2 (sin alpha_A cos alpha_A)/(sin alpha_B cos alpha_B) = (v_A/v_B)^2 (sin 2alpha_A)/(sin 2alpha_B)#

and the height ratio dictates the flight angle combinations allowed:

#(sin alpha_A)/(sin alpha_B) = sqrt(h_A/h_B)#,

meaning that the launch angle of #A# restricts the associated angles allowed for #B#.

There are an infinite number of launch angle combinations that satisfy the needed height ratio (provided #bb(0^@ < alpha_A < 90^@)# and the appropriate #alpha_B# is used), and that therefore influences the horizontal range.

I plotted the flight paths of #A# and #B# at #alpha_A = 30^@# and #alpha_A = 45^@# (with the appropriate angle for #alpha_B#), for three height ratios---#3:1#, #2:1#, and #1:1#.

Note that I used a speed ratio of #1/1#.

General observations:

• You can see the flight paths converge as the height ratios approach #1:1#.
• At a height ratio of #3:1#, the paths of #A# and #B# are generally distinct.
• At a height ratio of #2:1#, the #30^@# path of #A# is the #45^@# path of #B#, demonstrating how the height ratio restricts allowed #alpha_B#.

HEIGHT RATIO OF 3:1

The associated angle combinations can be found here.

HEIGHT RATIO OF 2:1

The associated angle combinations can be found here.

HEIGHT RATIO OF 1:1

The associated angle combinations can be found here.

Let the velocity of projection be #u# and angle of projection be #alpha#. #sin alpha# component of the initial velocity is responsible for reaching maximum height.
Let us use the following kinematic equation

#v^2 - u^2 = 2ah#

We know that at maximum height, #sin alpha# component is velocity #=0# also that gravity is acting against the direction of this velocity. We get

#0 - (usinalpha)^2 =- 2gh#

#=>h = ((usinalpha)^2)/(2g)#

Initial relation between the particles #Aand B# is set up as

#(h_A)/(h_B) = (((usinalpha_A)^2)/(2g))/(((usinalpha_B)^2)/(2g))#
#=>(h_A)/(h_B) = sin^2alpha_A/sin^2alpha_B= 3/1 = 3#

#=>(sinalpha_A)/(sinalpha_B) = sqrt3#

We also know that #"sine"# of any angle can not be greater than #1#.
Therefore, above equation #implies alpha_B# < #alpha_A#

To calculate horizontal range #R# we need to find out time of flight #T#. Using the kinematic expression
#h=ut+1/2g t^2# we get

#h = (usinalpha)t - 1/2g t^2#

For time of flight #h# is #=0#, we get

#0 = (usinalpha)T - 1/2g T^2#

#=>1/2g T = usinalpha#

#=>T = (2usinalpha)/g#

Range is given by the #cos alpha# component of initial velocity. Ignoring air resistance we have

#R = (ucosalpha)T#

Inserting value of #T# we get

#R = (ucosalpha)((2usinalpha)/g)#

#R = (2u^2sinalphacosalpha)/g#

This equations for particles #A and B# are

#R_A = (2u^2sinalpha_Acosalpha_A)/g#

#R_B = (2u^2sinalpha_Bcosalpha_B)/g#

Ratio of ranges after the speed of particle #A# is doubled can be written as

#(R'_A)/(R_B) = ((2(2u)^2sinalpha_Acosalpha_A)/g)/((2u^2sinalpha_Bcosalpha_B)/g)#

#=>(R'_A)/(R_B) = (4sin(2alpha_A))/(sin(2alpha_B))#

From practical point of view and a workable ratio #alpha_A# should be less than #90^@# and greater than #0^@# we have also shown that #alpha_B# < #alpha_A#

Let us assume #alpha_A=45^@#, angle for maximum range

we have #(sin45^@)/(sinalpha_B) = sqrt3#
#=>(1/sqrt2)/(sinalpha_B) = sqrt3#
#=>sinalpha_B = 1/sqrt6#
#=>alpha_B=24^@#

inserting in the ratio of ranges we get
#=>(R'_A)/(R_B) = 4/(sin(2xx24^@))#
#=>(R'_A)/R_B=4/0.7454=5.4#

As such we need additional information regarding one of the angles of projection to calculate numerical value of ratio.