A projectile is shot from the ground at an angle of #pi/4 # and a speed of #5 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

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Answer 1 · 2016-01-27T22:21:00

Find the time it takes to reach the maximum height, and use the x component of the displacement equation to find the distance travelled in that time.

First we break up the velocity into it's vector components at time 0, #v_x(0)# and #v_y(0)# by using the identities:

#v_x(0)=v(0)cos(theta)#
#v_y(0)=v(0)sin(theta)#

Since we're presumably ignoring air resistance, the acceleration components are

#a_x=0#
#a_y= -g#

So our functions for # v_x(t) # and #v_y(t)# becomes

#v_x(t)=v(0)cos(theta)#
#v_y(t)=v(0)sin(theta)-g*t#

We can find the time it takes to reach the maximum height by setting #v_y(t)=0# which gives:

#t=v(0)sin(theta)/g#

Then we use the displacement formula

#x(t)=x(0)+v_x t+1/2 a_xt^2#

Since #a_x=0#, and #x(0)=0# this simplifies to

#x(t)=v_x t#

Substitution of #v_x# yields

#x(t)=v(0)cos(theta) t#

and substituting for t yields:

#x(t)=(v(0)cos(theta)v(0)sin(theta))/g# which simplifies to

#x(t)=(v(0)^2cos(theta)sin(theta))/g# and further simplifies to

#x(t)=(v(0)^2sin(2theta))/(2g)#

This gives the range of the projectile at the maximum height.

It's interesting to note that this is exactly half the range equation for a projectile returning to the same height.

#R=(v(0)^2sin(2theta))/(g)#

which is derived using

#y(t)=y(0)+v_y t-1/2g t^2#

which is the displacement equation for y.