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.
