Question

An arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up be the positive direction. Because gravity is the force pulling the arrow down, the initial acceleration of the arrow is −9.8 meters per second squared.

Write the equation that represents this relationship
1. Transform the equation to the format that can most easily be used to find the maximum height. What is the height?
2. Transform the equation to a different format to determine how many seconds it would take for the arrow to hit the ground.
3. Show how you would use the equation in part (a) to determine approximately when the arrow would be 100 above the base of the cliff.
4. Show how you could determine the height of the arrow after 2 seconds?

Answer 1

H = 181.3m

t = 2.9s

V = 39.9 m/s

Explanation:

Given that:

Initial velocity V = 49 m/s

Acceleration g = - 9.8 m/s^2

Initial height h = 58.8 m

The equation that represents this relationship

V^2 = u^2 + 2g(h)

To find the maximum height. What is the height

At maximum height, V = 0

0 = 49^2 - 2 × 9.8 H

19.6H = 2401

H = 2401/19.6

H = 122.5 - h

H = 122.5 + 58.8

H = 181.3 m

How many seconds it would take for the arrow to hit the ground ?

h = ut + 1/2gt^2

181.3 = 49t + 0.5 × 9.8t^2

181.3 = 49t + 4.9t^2

Using quadratic formula

t = (-49 - 77.2)/ 9.8 or (-49+ 77.2)/9.8

t = positive

t = 2.9 s

Show how you would use the equation in part (a) to determine approximately when the arrow would be 100 above the base of the cliff.

.V^2 = u^2 + 2g*h.

V^2 = 0 + 19.6(181.3-100) = 1593.48, V = 39.9 m/s.

To determine the height of the arrow after 2 seconds

h = ut + 0.5gt^2.

U = 49 m/s, t = 2 s., g = -9.8 m/s^2, h h = 98 + 19.6

h = 78.4