Question

A projectile is fired from the edge of a cliff 200 m high with an initial speed of 30 m/s at an angle of 45° above the horizontal.

Determine the angle the velocity makes with the horizontal just before impact.

Answer 1

By the law of conservation of energy,

m(gh + 1/2u^2)= m(1/2v^2) ---------- cancelling m from both side

=> 4.8 * 200 + 1/2 * (30)^2 = 1/2(v^2)

=> 1460 + 450 = v^2/2

=> 2410 * 2 = v^2

=> v = √4820

= 69.426m/s

Answer: 69.426 m/s