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

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

Step-by-step 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 - (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) = (y_(f) - y_(0) + (1)/(2)gt^(2))/(t) = (0 - 9000 m + (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

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

Step-by-step 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 + (1)/(2) gt^2\\\\9000 = 90u + (1)/(2) (9.8)(90)^2\\\\9000 = 90u + 39690\\\\90u = -30690\\\\u = (-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* 9000)\\\\v^2 = 292681\\\\v = √(292681) \\\\v = 541 \ m/s`

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