A cannon ball is fired at an angle of 22 above horizontal with an initial speed

Question

A cannon ball is fired at an angle of 22 above horizontal with an initial speed v0 = 300 m/s, reaching a maximum height h before hitting the ground.

What is the maximum height h reached by the ball?

(a) 644 m
(b) 782 m
(c) 981 m
(d) 2420 m
(e) 3185 m

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What is the speed of the ball when it reaches its maximum height?
(a) 300 m/s
(b) 0 m/s
(c) 278 m/s

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Suppose the maximum height is h. Suppose now the cannon is fired at the same angle but with twice the initial speed (600 m/s). What will be the maximum height reached by the ball?
(a) 2 h
(b) 3 h
(c) 4 h

With the equation please

Explanation / Answer

h = v^2/2g =300sin 22 ^2/2*9.8 =644 m

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