A ball is launched from ground level at 28 m/s at an angle of 37° above the horizontal.How far does it go before it is at ground level again?

A) 58 m
B) 102 m
C) 38 m
D) 77 m
E) 19 m

Question

A ball is launched from ground level at 28 m/s at an angle of 37° above the horizontal.How far does it go before it is at ground level again?
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A) 58 m
B) 102 m
C) 38 m
D) 77 m
E) 19 m

Quizplus · Accepted Answer

Using the equations of motion, we can find the time taken for the ball to reach ground level again. 
Initial vertical velocity, u = 28 sin 37° = 16.8 m/s 
Time taken to reach the maximum height, t = u/g = 16.8/9.8 = 1.71 s 
Maximum height, h = ut - 1/2 gt^2 = 16.8(1.71) - 1/2 (9.8)(1.71)^2 = 12.3 m 
Time taken to reach ground level again = 2t = 3.42 s 
Horizontal distance covered = horizontal velocity x time = 28 cos 37° × 3.42 = 77 m

A Ball Is Launched from Ground Level at 28 M/s