Question

A projectile is shot at an angle of `pi/12` and a velocity of `6 m/s`. How far away will the projectile land?

Answer 1

`1.84"m"`

Explanation:

We 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:

`v=u+at`

This becomes:

`v_y=0-"g"t`

Since `v_y=vsintheta` this becomes:

`vsintheta=0-"g"t`

`:.t=(vsintheta)/g`

This is the time to reach maximum height so total time of flight `rArr`

`t_("tot")=(2vsintheta)/g" "color(red)((1))`

For the horizontal component:

`vcostheta=d/t_"tot"`

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

Putting `color(red)((1))` equal to `color(red)((2))rArr`

`{2vsintheta)/g=(d)/(vcostheta)`

`:.d=(v^(2)2sinthetacostheta)/g`

Since `sin2theta=2sinthetacostheta` this becomes:

`d=(v^2sin2theta)/g" "color(red)((3))`

Convert radians to degrees:

`pi=180^@`

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

`:.2theta=30^@`

`:.d=(6^2sin30)/9.8=(36xx0.5)/9.8=1.84"m"`

In future it is quicker if you can just remember equation `color(red)((3))`.