Question

A projectile shot at an angle of 60° above the horizontal strikes a building 30m away at a point 15m above the point of projection. Find the speed of projection.

Answer 1

Please verify for calculation errors. The procedure however is correct.

Answer 2

Answer:

Explanation:

use the formula y=xtanθ-gx2/(2u2cos2θ)

here y= 15m

x=30 m

θ=60

then calculate u

u=21.85 m/s (speed of projection)

magitude of the velocity when it hit the wall can be calculated as

v2=u2-2gh

=(21.85)2-2*9.81*15

v=13.53 m/s