Question

A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach the bottom of the cliff?What is its speed just before hitting?What total distance did it travel?

Answer 1

1)

vi = 15.0 m/s

d = - 75.0 m

a = -9.8 m/s^2

use:

d = vi*t + 0.5*a*t^2

-75 = 15.0*t + 0.5*(-9.8)*t^2

-75 = 15*t - 4.9*t^2

4.9*t^2 -15t - 75 = 0

This is quadratic equation (at^2+bt+c=0)

a = 4.9

b = -15

c = -75

Roots can be found by

t = {-b + sqrt(b^2-4*a*c)}/2a

t = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.695*10^3

roots are :

t = 5.732 and t = -2.67

since t can't be negative, the possible value of t is

t = 5.73 s

Answer: 5.73 s

2)

vf^2 = vi^2 + 2*a*d

vf^2 = 15.0^2 + 2*(-9.8)*(-75.0)

= 225 + 1470

vf^2 = 1695

vf = 41.2 m/s

Answer: 41.2 m/s

3)

Lets calculate the height reached from cliff

vf = 0 m/s (At maximum height, velocity is 0)

vi = 15.0 m/s

a = -9.8 m/s^2

Use:

vf^2 = vi^2 + 2*a*d

0 = 15.0^2 + 2*(-9.8)*d

19.6*d = 225

d = 11.5 m

So, the stone goes 11.5 up, comes 11.5 down and finally goes 75.0 m further down

Total distance = 11.5 + 11.5 + 75.0

= 98.0 m

Answer: 98.0 m