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Maksim231197 [3]
4 years ago
6

A student standing on the ground next to a tall building throws a ball straight up with a speed of 39.2 m/s. If it takes the bal

l 4 seconds to reach the same height as the top of the building, how tall is the building?
Physics
1 answer:
cricket20 [7]4 years ago
6 0
Using the equation of motion:

h = ut - 0.5gt²,  

 g  is -ve for a rising body,     g ≈ 10 m/s²,  t = 4s, u = 39.2m/s

Note that in this problem we are neglecting the height of the student.

h = ut - 0.5gt²,

h = 39.2*4<span> - 0.5*10*4²            Use a calculator
</span>
h = 76.8 m

The building is 76.8 m tall.
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An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is vi and second
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Answer:

The maximum height reached in the second trial is 16times the maximum height reached in the first trial.

Explanation:

The following data were obtained from the question:

First trial

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Final speed (v) = 0

Second trial

Initial speed (u) = 4v

Final speed (v) = 0

Next, we shall obtain the expression for the maximum height reached in each case.

This is illustrated below:

First trial:

Initial speed (u) = v

Final speed (v) = 0

Acceleration due to gravity (g) = 9.8 m/s²

Height (h₁) =.?

v² = u² – 2gh₁ (going against gravity)

0 = (v)² – 2 × 9.8 × h₁

0 = v² – 19.6 × h₁

Rearrange

19.6 × h₁ = v²

Divide both side by 19.6

h₁ = v²/19.6

Second trial

Initial speed (u) = 4v

Final speed (v) = 0

Acceleration due to gravity (g) = 9.8 m/s²

Height (h₂) =.?

v² = u² – 2gh₂ (going against gravity)

0 = (4v)² – 2 × 9.8 × h₂

0 = 16v² – 19.6 × h₂

Rearrange

19.6 × h₂ =16v²

Divide both side by 19.6

h₂ = 16v²/19.6

Now, we shall determine the ratio of the maximum height reached in the second trial to that of the first trial.

This is illustrated below:

Second trial:

h₂ = 16v²/19.6

First trial:

h₁ = v²/19.6

Second trial : First trial

h₂ : h₁

h₂ / h₁ = 16v²/19.6 ÷ v²/19.6

h₂ / h₁ = 16v²/19.6 × 19.6/v²

h₂ / h₁ = 16

h₂ = 16 × h₁

From the above illustrations, we can see that the maximum height reached in the second trial is 16times the maximum height reached in the first trial.

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