A stone is thrown vertically upward with a speed of 14.3 m/s from the edge of a

Question

A stone is thrown vertically upward with a speed of 14.3 m/s from the edge of a cliff 75.0 m high (Figure 1).

Part A

How much later does it reach the bottom of the cliff?

Express your answer to three significant figures and include the appropriate units.

Part B

What is its speed just before hitting?

Express your answer to three significant figures and include the appropriate units.

Part C

What total distance did it travel?

Express your answer to three significant figures and include the appropriate units.

Explanation / Answer

A)

d = -75.0 m

a = -9.8 m/s^2

vi = 14.3 m/s

use:

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

-75.0 = 14.3*t + 0.5*(-9.8)*t^2

-75.0 = 14.3*t - 4.9*t^2

4.9*t^2 - 14.3*t - 75.0 = 0

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

a = 4.9

b = -14.3

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 = 1.674*10^3

roots are :

t = 5.635 and t = -2.716

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

t = 5.64

Answer: 5.64

B)

a = -9.8 m/s^2

vi = 14.3 m/s

t= 5.64 s

use:

vf = vi + a*t

= 14.3 - 9.8*5.64

= -41.0 m/s

Answer: -41.0 m/s

C)

Lets calculate the maximum height

vi = 14.3 m/s

vf = 0 m/s

a = -9.8 m/s^2

use:

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

0 = 14.3^2 + 2*(-9.8)*d

19.6*d = 204.49

d = 10.4 m

So, it goes 10.4 m up, comes 10.4 m and again fall 75.0 m

Total distance = 10.4 + 10.4 + 75.0

= 95.8 m

Answer: 95.8 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.