Question

A golfer hits a shot to a green that is elevated 2.80 m above the point where the ball is struck. The ball leaves the club at a speed of 18.7 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:

32.812m/s

Explanation:

Now the time of the projectile motion is given by;

t = usinA/ g

Where A is angle =49°

u is initial velocity,u = 18.7m/s

Hence t = 18.7 ×sin49°/ 9.8 = 1.44s

The final velocity from Newton's law V = U + gt

= 18.7 + (9.8 ×1.44)= 32.812m/s