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Leona [35]
3 years ago
10

A student throws a heavy ball downward off the top of a building with a speed of 18m/s. The ball reaches a speed of 41m/s just b

efore striking the ground. Neglect drag.
Mathematics
1 answer:
DedPeter [7]3 years ago
4 0

Complete question:

A student throws a heavy ball downward off the top of a building with a speed of 18m/s. The ball reaches a speed of 41m/s just before striking the ground. Neglect drag, find the height of the building.

Answer:

The height of the building is 69.235 m

Step-by-step explanation:

Given;

initial velocity of the ball, u = 18 m/s

final velocity of the ball, v = 41 m/s

The height of the building is equal to distance traveled by the ball downward.

Apply the following kinematic equation;

v² = u² + 2gh

where;

g is acceleration due to gravity

h is height of the building

41² = 18² + 2(9.8)h

1681 = 324 + 19.6h

19.6h = 1681 - 324

19.6h = 1357

h = 1357 / 19.6

h = 69.235 m

Therefore, the height of the building is 69.235 m

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