Question

The acceleration due to gravity is 9.81 m/s2, towards the Earth. Rain falling from an altitude of 9,000 m would fall for about 1.5 minutes before hitting the ground (ignoring air resistance). What would be the final speed of a raindrop by the time it reaches the ground?

Answer 1

Answer:

The magnitude of the final speed of a raindrop by the time it reaches the ground will be 541.45 m/s.

Explanation:

The final speed of the raindrop can be found using the following equation:

$v_{f} = v_{0} - gt$

Where:

$v_{f}$: is the final speed =?

$v_{0}$: is the initial speed

g: is the acceleration due to gravity = 9.81 m/s²

t: is the time = 1.5 min

First, we need to find the initial speed:

$y_{f} = y_{0} + v_{0}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$: is the final height = 0

$y_{0}$: is the initial height = 9000 m  

Hence, the initial speed is:    

$v_{0} = \frac{y_{f} - y_{0} + \frac{1}{2}gt^{2}}{t} = \frac{0 - 9000 m + \frac{1}{2}9.81 m/s^{2}*(90 s)^{2}}{90 s} = 341.45 m/s$

Hence, the final speed is:                            

$v_{f} = v_{0} - gt = 341.45 m/s - 9.81 m/s^{2}*90 s = -541.45 m/s$

Therefore, the magnitude of the final speed of a raindrop by the time it reaches the ground will be 541.45 m/s.

           

I hope it helps you!                      

Answer 2

Answer:

the final speed of the rain is 541 m/s.

Explanation:

Given;

acceleration due to gravity, g = 9.81 m/s²

height of fall of the rain, h = 9,000 m

time of the rain fall, t = 1.5 minutes = 90 s

Determine the initial velocity of the rain, as follows;

$h = ut + \frac{1}{2} gt^2\\\\9000 = 90u + \frac{1}{2} (9.8)(90)^2\\\\9000 = 90u + 39690\\\\90u = -30690\\\\u = \frac{-30690}{90} \\\\u = -341 \ m/s$

The final speed of the rain is calculated as;

$v^2 = u^2 + 2gh\\\\v^2 = (-341)^2 + 2(9.8\times 9000)\\\\v^2 = 292681\\\\v = \sqrt{292681} \\\\v = 541 \ m/s$

Therefore, the final speed of the rain is 541 m/s.