Question

If a golf ball is hit with an initial velocity of 81.4 m/s at an angle of 18.9 what would be the amount of time that the ball spends in the air before it hits the ground (returning to the same height from which it was hit)?

Answer 1

Given that initial speed of ball, u=81.4 m/s making an angle, x = 18.9 degree

The total time taken will be,

`t=\frac{2u\sin \text{ x}}{g}`

Here, g is acceleration due to gravity whose value is 9.8 m/s.

Substituting the values, we get

`t=\frac{2*81.4*\sin (18.9^(\circ)\text{)}}{9.8}`

Hence the total time is 5.38 seconds.