Question

If a projectile is shot at a velocity of `2 m/s` and an angle of `pi/4`, how far will the projectile travel before landing?

Answer 1

it will land about 0.638 m far.

Explanation:

We have to write down the equations of motion:

`x= x_0+vcosthetat`
`y= h_0+ v sintheta t + (at^2)/2`

where `v` is the velocity of the projectile, `theta` is the angle of the trajectory, `h_0` the initial quota of the projectile, `x_0` its initial position.

We have:

• `h_0=0` and `x_0=0`;
• `v=2 \ ms^(-1)`
• `theta= pi/4`
• `a=-g \ ~~ -9.81 \ ms^(-2).`

We have `v` and `thetha`: we have to find t from the first equation and substitute its value in the second one.
When the projectile lands, it reaches the ground so we have y=0 at that moment.
The equations of motion at the landing are:

`x= vcosthetat`
`0= v sintheta t - (gt^2)/2`

from the first equation `t=x/(vcostheta)`.
The other becomes:
`v sin theta * x/(vcostheta) - g/2*(x/(vcostheta))^2=`

Performing the calculation we eventually have:
`x= sqrt( (2v^2 costheta sintheta)/g )~~ 0.638 \ m`