Question

6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above the tee. A golfer hits a ball at an angle of 8.6° above the horizontal and makes a hole-in-one! What was the initial speed of the ball?

Answer 1

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of  12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s