Question

A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck. The ball leaves the club at a speed of 16.6 m/s at an angle of 40.9° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.

Answer 1

Answer:14.72 m/s

Step-by-step explanation:

Given

Initial velocity (u)=16.6 m/s

`\theta =40.9^(\circ)`

Horizontal velocity component (
`u_x`)=16.6cos40.9=12.54 m/s

As the ball comes down so its vertical displacement is zero except 3 m elevation

Thus
`v_y=√(\left ( 16.6sin40.9\right )^2+2\left ( -9.81\right )\left ( 3\right ))`

`v_y=√(10.868^2-58.86)`

`v_y=√(59.253)`

`v_y=7.69 m/s`

there will be no change is horizontal velocity as there is no acceleration

Therefore Final Velocity

`v=√(u_x^2+v_y^2)`

`v=√(12.54^2+7.69^2)`

v=14.72 m/s