Ah yes the language of enchantment book
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.
Answer:
The fireman continues to descend but with constant speed.
Explanation:
According to Newton first law of motion, The object which is rest continues to remain at rest or object moving with constant speed move with constant velocity when acted upon by an unbalanced force.
When the net force acting on the object becomes zero the object moves with constant speed.
Hence when the frictional force acting on the fireman hands equals to his body weight , the net force acting on the fireman becomes zero and thus fireman continues to descend with constant speed.