Question

A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 18.1 m/s at an angle of 49.0° 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

16.17 m/s

Step-by-step explanation:

h = 3.2 m

u = 18.1 m/s

Angle of projection, θ = 49°

Let H be the maximum height reached by the ball.

The formula for the maximum height is given by

`H=(u^(2)Sin^(2)\theta )/(2g)`

`H=(18.1^(2)* Sin^(2)49 )/(2* 9.8)=9.52 m`

The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m

Let v be the velocity of ball with which it strikes the ground.

Use third equation of motion for vertical direction

`v_(y)^(2)=u_(y)^(2)+2gh'`

here, uy = 0

So,

`v_(y)^(2)=2* 9.8 * 6.32`

vy = 11.13 m/s

vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s

The resultant velocity is given by

`v=\sqrt{v_(x)^(2)+v_(y)^(2)}`

`v=\sqrt{11.87^(2)+11.13^(2)}`

v = 16.27 m/s