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telo118 [61]
3 years ago
14

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?
Mathematics
1 answer:
Natali5045456 [20]3 years ago
4 0

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

Step-by-step 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=\frac{V-U}{a}

   t=\frac{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

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