Question

A juggler tosses a bowling pin in the air with an initial velocity of 18 ft per second. It leaves his hand when it is five feet from the ground and he catches it when it is four feet from the ground?

Answer 1

Complete Question

A juggler tosses a bowling pin in the air with an initial velocity of 18 ft per second. It leaves his hand when it is five feet from the ground and he catches it when it is four feet from the ground,How long is the ball in the air?

Answer:

`t\approx 0.1s`

Explanation:

From the question we are told that

Initial velocity
`V_0= 18 ft/s`

Initial Height
`h_1=5feet`

Final Height
`h_2=4feet`

Generally the Newtons equation for motion is mathematically given by

`V^2 = U^2 + 2as`

Where

`s=h_2-h_1\\s=4-5\\s=-1ft`

And

`a=g=>9.81\\a=32.2ft/s^2`

Therefore

`V^2 = (18)^2 + 2(-32.2)(-1)`

`V^2 = 388.4`

`V =19.70786645\\V=19.71ft/s`

Generally the equation for Time is mathematically Given as

`t=(V-U)/(a)`

`t=(19.71-18)/(32.2)`

`t=0.05310559006`

`t\approx 0.1s`

Therefore the time spent on air by the ball is

`t=0.05310559006s`