Answer:
The velocity of the droplet right before it hits the ground is 40.08 m/s.
Explanation:
To determine the velocity of the droplet right before it hits the ground,
From one of the equations of kinematic for free fall motions,
v = u + gt
Where v is the final velocity
u is the initial velocity
g is acceleration due to gravity (take g = 9.8 m/s²)
and t is time
For the question, v is the velocity of the droplet right before it hits the ground.
u = 0 m/s (Since the molten lead was dropped from rest)
Therefore,
v = gt
First, we will determine the time t
Also, from one of the equations of kinematic for free fall motions,
h = ut + 1/2(gt²)
u = 0 m/s
From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore
h = 82.15
Hence,
82.15 = 0×t + 1/2 (9.8 × t²)
82.15 = 1/2 (9.8 × t²)
82.15 = 4.9 t²
t² = 82.15/4.9
∴ t = 4.09 secs
Now, for the velocity v, of the droplet right before it hits the ground,
Recall
v = gt
Then,
v = 9.8 × 4.09
v = 40.08 m/s
Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.
