Q73P - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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.

Step1: Identification of given data

Acceleration  $a=35m/s^2$

Distance  $d=64 cm=0.64m$

Step 2: Calculation for the speed

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

  $$\begin{gathered} s=\frac{d}{t} \\ s=\frac{0.64}{0.191} \\ s=3.35m/s \end{gathered}$$

Step 3: Calculation for the height above the ground it goes

Height above the ground it goes is 1.83 m.

Step 4: Calculation for the time required

$$\begin{gathered} d=ut+\frac{1}{2}a{t}^2 \\ 0.64=0+\frac{1}{2}\times 35\times t^2 \\ t=0.191 s \end{gathered}$$