Question

A 3.8 lb object is thrown vertically upward with an initial velocity of 36.0 m/s. What maximum height does it reach?   It now falls vertically from this height. What is its velocity just before hitting the ground?  When it came to rest, it left a 1.30 cm indentation in the ground. What was the magnitude and direction of the force that brought the object to rest?

Answer 1

Mass = 3.8 lb = 1.724 kg

Initial Velocity =36 m/s

acceleration =-g

Max Height =?

Use formula v² -u² =2as

At max height final velocity is ZERO

0- 36² = -2 *9.81 *H

H = 36²/2*9.81 = 66.06 m

ANSWER: Max height reached = 66.06 m

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Now the falling part

Initial velocity is ZERO ,find final velocity just before hitting the ground

Use formula v² -u² =2as

V² - 0 = 2 *9.81 *H

V =2*9.81*66.06 = 36m/s

Answer : V =36m/s

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When a object falls from a height H .

Workdone = mgh = 1.724 kg *9.81* H

= 1.724 *9.81*66.06 = 1117.24 J

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Now this energy is lost to ground on hitting the ground and making an indentation of 1.3 cm

W =Force * distance

1117.24 J = F * 0.013

F = 1117.24/0.013

=85941.5 N

Force will be acting opposite to the direction of falling object

that is upwards

ANSER : F =85941.5 N (upwards)