Question

Part c

Chapter 2 Problem 2.73 Part A Constants Sam heaves a shct with weight 16-lb straight upward, giving it a constant upward acceleration from rest of 347 m/st for a height 69.0 cm. He releases it at height 2.12 m above the ground. You may ignore air resistance. What is the speed of the shot when he releases it? 6.92 m/s Previous Answers Correct Part B How high above the ground does it go? h 4.56 m Correct Part C How much time does he have to get out of its way belore it returns to the height of the top of his t 1.3938 Submit Previous Answers Request Answer X Incorrect, Try Again; 4 attempts remaining

Answer 1

Given : W = 66 lb , a = 34.7 m/s², d= 69 cm (0.69 m) , h₁= 2.12 m

Solution:

(a) Speed of shot when he releases it

As the acceleration is constant , applying equation of motion to get velocity:

v² = v₀² + 2ad

= 0 + 2(34.7)(0.69)

= 47.886

i.e. $v = \sqrt 47.886$ = 6.92 m/s

Answer : v = 6.92 m/s

(b) height above the ground

We know that, velocity at highest point will be zero.

Applying equation of motion :

v² = v₀² + 2ah₂

Here , a = -g = -9.81 m/s² and v = 0

0 = (6.92)² - 2(9.81)h₂

i.e.  $h_2 = \frac{47.886}{19.62}$ = 2.44 m

Now the height it reaches above ground is given by:

h = h₁ + h₂ = 2.12 m + 2.44 m = 4.56 m

Answer : h = 4.56 m

(c) the time that he have to get out of its way before it returns to the height of the top of his head

Using equation of motion , d = v₀t + (1/2)at²

(4.56 - h) = 0 + (1/2)(9.81)t²

$t = \sqrt \frac{(4.56-h}{4.905}$

where h is the height of head from ground