Question

15. Lee hit a golf ball with an initial velocity of 150 feet per second at an angle of 45 degrees above the horizontal. Calculate how long the ball will be in the air to the nearest tenth of a second.

Answer 1

Answer:
`21.6\ s`

Explanation:

Given

The initial velocity of the ball is
`u=150\ ft\s`

Angle of projection
`\theta=45^(\circ)`

The time of flight of the ball is given by

`T=(2u\sin \theta)/(g)\\\\`

Insert the values

`T=(2* 150* \sin 45^(\circ))/(9.8)\\\\T=(212.132)/(9.8)\\\\T=21.6\ s`