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Nitella
14 days ago
8

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?
Mathematics
1 answer:
lawyer [12.5K]14 days ago
7 0

The result is:

H = 181.3m

t = 2.9s

Here’s the breakdown of the explanation:

Provided information:

Initial speed V = 49 m/s

Acceleration g = - 9.8 m/s^2

Starting height h = 58.8 m

The equation that captures this scenario

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

To ascertain the maximum height. What's the height

At the peak, V = 0

0 = 49^2 - 2 × 9.8 H

19.6H = 2401

From here, H = 2401/19.6

H = 122.5 - h

Resulting in H = 122.5 + 58.8

Therefore, H = 181.3 m

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

h = ut + 1/2gt^2

181.3 = 49t + 0.5 × 9.8t^2

181.3 = 49t + 4.9t^2

Utilizing the quadratic formula

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

t = positive

t = 2.9 s

To determine the time when the arrow is approximately 100 up from the cliff’s base using the earlier equation.

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

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

To evaluate 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

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