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

u = 104.68 m/s

Step-by-step 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 - (1)/(2)gt^2`................(2)

from equation(1) and (2)

`y = x tan \theta - (gx^2)/(2u^2cos^2\theta)`..........{3}

`12.4 = 150* tan (8.6) - (9.8* 150^2)/(2u^2cos^2(8.6))`

`(9.8* 150^2)/(2u^2cos^2(8.6)) = 10.29`

`(9.8* 150^2)/(2* 10.29* cos^2(8.6)) = u^2`

`u = √(10959.34)`

u = 104.68 m/s

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