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

Answer:

16.17 m/s

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

$H=\frac{u^{2}Sin^{2}\theta }{2g}$

$H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m$

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,

$v_{y}^{2}=2\times 9.8 \times 6.32$

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