Question

A ball is thrown straight up with an initial speed of 19 m/s. The ball was released 6.3 m above the ground, but when it returns back down, it falls into a hole d m deep. If the ball’s speed is 35.7 m/s at the bottom of the hole, how deep is the hole (in m)?

Answer 1

When the ball is thrown from a height of 6.3 m it will go upto a maximum height where its velocity will become 0 means final velocity will be 0 .

Initial velocity , u = 19 m/s

Final velocity, v = 0 m/s

So the distance covered up to maximum height from 6.3 m.

By using third equation of motion.

v²   = u² + 2.a.s

Where , a is acceleration due to gravity , a= -g , negative g because ball is going above and acceleration due to gravity is downward.

Take g = 9.8 m/s2

s is the distance covered.

0 = 19*19 - 2*9.8*s

s = 18.418 m

When ball will return from maximum height now this time initial velocity will be 0 because already at maximum height velocity is 0, and final velocity is 35.7 m/s

Again by using third equation of motion.

v²   = u² + 2.a.s

35.7*35.7 = 0 + 2*9.8*s'

s' =65.025 m

Total returning path ( s' ) = 6.3 + s + d

Where s is height from 6.3 to maximum height, d is depth of the hole

6.3 + 18.418 + d = 65.025

d = 40.307 m