Question

3) A cannon is launched upward at a 42 degree angle with a speed of 100 m/s from a height of 50 m. a) How long does it take for the cannon ball take to hit the ground (0 m)? b) What is the range of the cannon?

Answer 1

A)

Using 2nd equation of motion

h = ho + v sin x t - 0.5 gt^2

0 = 50 + 100 sin 42 * t - 4.9 t^2

Solving for t

t = 14.37 sec

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B)

Range of the cannon

R = v cos x * t

R = 100 cos 42 * 14.37

R = 1067.90 m

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