Question

A projectile is shot from the ground at an angle of `pi/8` and a speed of `5 /7 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?

Answer 1

0.020m

Explanation:

There are two ways to solve this problem. First, the long way.
We first find the vertical and horizontal velocities which are

`u_H=ucostheta=0.660ms^-1`
`u_V=usintheta=0.273ms^-1`

Next, we must use the equation

`v=u+at`

to find the time it took for the projectile to reach its maximum height.
We first rearrage to find t, which is

`(v-u)/a=t`

v in this case would be zero, as when the ball reaches it's maximum height, it has 0 vertical velocity. a would be acceleration due to gravity and u would be the value we calculated to initial vertical velocity. Thus we get

`((0-(0.273))/-9.81)=0.0278`seconds.

This is the time it took for the projectile to reach it's maximum height. We now can find the distance, vertically and horizontally the projectile will travel.

The vertical distance it will travel when reaching it maximum height can be found using

`v^2=u^2+as`

Rearragned, it gives us distance

`s=(v^2-u^2)/a`

Substituting the values in to get the verticaldistance, we get

`s=(0^2-(0.273)^2)/-9.81=0.0076`

Now to get the horizontal distance, we simply use

`s=vt`

whcih gives us

`s=(0.660)(0.0278)=0.018`

Using pythagoras, we get the total final distance which would be

`s=sqrt(0.018^2+0.0076^2)=0.020`