Question

Using your trusty sling-shot, you fire a 3.00-N rock vertically into the air. You observe that when the rock is 15.5 m above where you launched it, it is traveling at 24.9 m/s upward.

A. Use energy concepts to determine the launch speed of the rock.

B. Use energy concepts to find the rock's maximum height.

Answer 1

Given, Weight of rock (W) = 3 N , h = 15.5 m , v = 24.9 m/s

So, mass of the rock (m) = 3/9.8 = 0.306 kg

a) Using conservation of energy ,

(1/2)*m*u^2 = mgh + (1/2)*mv^2

(1/2)*0.306*u^2 = (0.306*9.8*15.5) + (1/2)*0.306*24.9^2

(1/2)*0.306*u^2 = 141.343

u^2 = 923.81

u = 30.394 m/s

Thus , the launch speed of the rocket is 30.394 m/s

b) Using conservation of energy ,

(1/2)*mu^2 = mg*h1

(1/2)*0.306*30.394^2 = 0.306*9.8*h1

h1 = 141.34/(0.306*9.8)

h1 = 47.132 m

Thus , maximum height is 47.132 m