Question

6 A small rock is thrown vertically upward with a speed of 14.0 m/s from the edge of the roof of a 30.0 m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.  

A) What is the speed of the rock just before it hits the street?
B)

How much time elapses from when the rock is thrown until it hits the street?

Answer 1

Refer to the diagram shown below.

Air resistance is ignored, and g = 9.8 m/s².
All quantities are measured as positive upward.

Part A)
The ground is at a distance of s = - 30 m from the launch point.
If the rock strikes the ground with velocity,v, then
v² = u² + 2as
v² = (14 m/s)² - 2*(9.8 m/s²)(-30 m) = 784 (m/s)²
= 28 m/s

Answer: 28 m/s

Part B)
Let t = time for the rock to hit the ground.
s = ut + (1/2)at²
(-30 m) = (14 m/s)*(t s) + 0.5*(-9.8 m/s²)*(t s)²
-30 = 14t - 4.9t²
4.9t² - 14t - 30 = 0
t² - 2.857t - 6.1224 = 0

Solve with the quadratic formula.
t = (1/2) [2.8571 +/- √(8.163+24.4896)]
= 4.286 or -1.429 s
Reject negative time.

Answer: 4.29 s (nearest hundredth)