Question

A projectile is shot from the ground at a velocity of `5 m/s` and at an angle of `(7pi)/12`. How long will it take for the projectile to land?

Answer 1

`0,9856s`

Explanation:

The only force acting on the projectile after being fired is the force of gravity.

We may resolve the motion into 2 components :

In the horizontal (X) direction, there is constant velocity of `5cos(7pi)/12rad=5cos105^@=5cos75^@1,2941m//2`

Note that we take the angle with the horizontal axis as the reference angle and so the projectile is actually being fired backwards with a projection angle of `75^@`.

In the vertical (Y) direction, there is constant acceleration due to gravity and hence the equations of motion for uniform acceleration in 1 direction are valid.

The initial velocity in the y-direction is hence `5sin75^@=4,8296m//s`.

Therefore, time taken to reach maximum height is given by :
`v=u+at => t=(v-u)/a`
`therefore t=(0-4,8296)/(-9,8)=0,4928s`.

Therefore the total time taken to reach ground again, since the path is symmetric, is `0,4928xx2=0,9856s`.