Question

A rock is thrown vertically upward with a speed of 14.0 m/sm/s from the roof of a building that is 70.0 mm above the ground. Assume free fall.A) In how many seconds after being thrown does the rock strike the ground? B) What is the speed of the rock just before it strikes the ground?

Answer 1

Answer:

(A). The time is 5.47 sec.

(B). The speed of the rock just before it strikes the ground is 39.59 m/s.

Explanation:

Given that,

Initial velocity = 14.0 m/s

Height = 70.0 m

(A). We need to calculate the time

Using second equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$70=-14\times t+\dfrac{1}{2}\times9.8\times t^2$

$4.9t^2-14t-70=0$

$t =5.47\ sec$

(B). We need to calculate the speed of the rock just before it strikes the ground

Using third equation of motion

$v^2=u^2+2gs$

Put the value into the formula

$v^2=(14)^2+2\times9.8\times70$

$v^2=1568$

$v=\sqrt{1568}$

$v=39.59\ m/s$

Hence, (A). The time is 5.47 sec.

(B). The speed of the rock just before it strikes the ground is 39.59 m/s.