Question

a man stands on the roof of a 18.0 m - tall building and throws a rock with a velocity of magnitude 30.0 m/s at an angle of 42.0° above the horizontal. you can ignore air resistance. part a calculate the maximum height above the roof reached by the rock. part b calculate the magnitude of the velocity of the rock just before it strikes the ground.

Explanation:

Part A: Max height above roof

Step1: Vertical initial velocity

\( v_{0y} = 30.0 \times \sin(40.0^\circ) \approx 19.28 \, \text{m/s} \)

Step2: Max height formula

\( h_{\text{max}} = \frac{v_{0y}^2}{2g} = \frac{(19.28)^2}{2 \times 9.8} \approx 19.0 \, \text{m} \)

Part B: Velocity before ground

Step1: Building height in meters

\( h = 16.0 \, \text{ft} \times 0.3048 \approx 4.877 \, \text{m} \)

Step2: Velocity magnitude formula

\( v = \sqrt{v_0^2 + 2gh} = \sqrt{30.0^2 + 2 \times 9.8 \times 4.877} \approx 31.6 \, \text{m/s} \)

Answer:

Part A: \( h_{\text{max}} = 19.0 \, \text{m} \)
Part B: \( v = 31.6 \, \text{m/s} \)