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Chapter 1: Q73P (page 63)

Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s2 for 64.0 cm. He releases it 2.20 m above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?

Short Answer

Expert verified

a) Speed of the shot is 3.35 m/s.

b) Height above the ground it goes is 1.83 m.

c) Time required is 0.191 s.

Step by step solution

01

Identification of given data

Acceleration a=35m/s2

Distance d=64cm=0.64m

02

Calculation for the speed

As we know that speed is the distance covered per unit time

s=dts=0.640.191s=3.35m/s

03

Calculation for the height above the ground it goes

Height above the ground it goes is 1.83 m.

04

Calculation for the time required

d=ut+12at20.64=0+12脳35脳t2t=0.191s

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