A ball is launched from a 300m cliff and lands 380m away from the cliff in 9 seconds. Calculate the initual speed and the angle of the ball when it was launched.

Explanation:

Given

Maximum height H = 300m

Range (horizontal distance) = 380m

Required

Initial speed U and the angle of the ball when it was launched.

Range = U√2H/g

380 = U√2(300)/9.8

380 = U√600/9.8

380 = 7.8246U

U = 380/7.8246

U = 48.57m/s

The initial speed is 48.57m/s

b) Using the formula for calculating time of flight;

T = 2Usin theta/g

9 = 2(48.57)sin theta/9.8

9*9.8 = 97.14sin theta

88.2 = 97.14sin theta

88.2/97.14 = sin theta

sin theta = 0.9079

theta = sin^-1(0.9079)

theta = 65.23°

hence the angle when the ball was launched is 65.23°

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