Question

a rock is thrown vertically upward with a speed of 12.0 m/s from the roof of a building that is 70.0 m above the ground. assume free - fall.

part a
in how many seconds after being thrown does the rock strike the ground?

part b
what is the speed of the rock just before it strikes the ground?

Explanation:

Step1: Set up displacement equation

\( y(t) = y_0 + v_0 t + \frac{1}{2} a t^2 \), where \( y=0 \), \( y_0=70.0 \, \text{m} \), \( v_0=12.0 \, \text{m/s} \), \( a=-9.8 \, \text{m/s}^2 \)

Step2: Substitute values

\( 0 = 70.0 + 12.0t - 4.9t^2 \) → \( 4.9t^2 - 12.0t - 70.0 = 0 \)

Step3: Solve quadratic equation

\( t = \frac{12.0 + \sqrt{(-12.0)^2 - 4×4.9×(-70.0)}}{2×4.9} \approx 5.20 \, \text{s} \)

Step4: Set up velocity equation

\( v^2 = v_0^2 + 2a(y - y_0) \)

Step5: Substitute values for velocity

\( v^2 = 12.0^2 + 2×(-9.8)(-70.0) = 1516 \) → \( v \approx 38.9 \, \text{m/s} \)

Answer:

5.20 seconds
38.9 m/s