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?

1 Answer
Jan 27, 2016

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.

Explanation:

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.