Question

A projectile is shot from the ground at an angle of `pi/12` and a speed of `7 /8 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

`0.0195"m"`

Explanation:

The key to these projectile problems is to treat the horizontal and vertical components of motion separately and use the fact that they share the same time of flight.

For the vertical component of motion we use:

`v=u+at`

This becomes:

`0=vsintheta-"g"t`

`:.t=(vsintheta)/g" "color(red)((1))`

Now we find `t` using the horizontal component:

In this case there is no acceleration on the particle so we can write:

`vcostheta=d/t`

`:.t=d/(vcostheta)" "color(red)((2))`

Now we can put `color(red)((1))` equal to `color(red)((2))rArr`

`(vsintheta)/g=d/(vcostheta)`

`:.d=(v^(2)sinthetacostheta)/g" "color(red)((3))`

We can simplify this using the trig identity:

`sin2theta=2sinthetacostheta`

So `color(red)((3))` becomes`rArr`

`d=(v^2sin2theta)/(2g)`

(nb this is the distance to reach maximum height which is 1/2 the total distance travelled. If you double this you get a handy expression for the range of the object.)

Converting `theta` into degrees:

`pi=180^@`

`:.pi/12=180/12=15^@=theta`

`:.2theta=30^@`

`v=7/8=0.875"m/s"`

Now put in the numbers `rArr`

`d=(0.875^(2)xx0.5)/(2xx9.8)`

`d=0.0195"m"`