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?

1 Answer
Dec 25, 2015

Answer:

#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#.