Question

An archer practicing with an arrow bow shoots an arrow straight up two times. The first time the initial speed is vi and second

Answer 1

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

Initial speed (u) = v

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.