A rugby player passes the ball 7.25 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used

Question

Hillary Daniel

Physics

1 July, 07:06

A rugby player passes the ball 7.25 m across the field, where it is caught at the same height as it left his hand. at what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used

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Answers (1)

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Adalynn Mccoy · Answer 1

Let x = angle at which the ball was thrown, relative to the horizontal.

The vertical component of velocity is

Vy = (12 m/s) sin (x) = 12sin (x) m/s

At maximum height, the vertical velocity is zero. The time to attain maximum height is given by

12sin (x) - gt = 0

t = [12sin (x) ]/9.8 = 1.2245sin (x) s

The horizontal component of velocity is 12cos (x) m/s.

The time taken to travel 7.25 m is 2t. Because distance = velocity*time, therefore

12cos (x) * (2*1.2245sin (x)) = 7.25

14.694 * (2sin (x) cos (x)) = 7.25

Note that 2sin (x) cos (x) = sin (2x). Therefore

sin (2x) = 7.25/14.694 = 0.4934

2x = arcsin (0.4934) = 29.56°

x = 14.8° (nearest tenth)

Answer: 14.8° (approx. 15°)