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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;
[tex]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[/tex]
The final speed of the rain is calculated as;
[tex]v^2 = u^2 + 2gh\\\\v^2 = (-341)^2 + 2(9.8\times 9000)\\\\v^2 = 292681\\\\v = \sqrt{292681} \\\\v = 541 \ m/s[/tex]
Therefore, the final speed of the rain is 541 m/s.
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:
[tex] v_{f} = v_{0} - gt [/tex]
Where:
[tex]v_{f}[/tex]: is the final speed =?
[tex]v_{0}[/tex]: 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:
[tex] y_{f} = y_{0} + v_{0}t - \frac{1}{2}gt^{2} [/tex]
Where:
[tex]y_{f}[/tex]: is the final height = 0
[tex]y_{0}[/tex]: is the initial height = 9000 m
Hence, the initial speed is:
[tex] 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 [/tex]
Hence, the final speed is:
[tex] v_{f} = v_{0} - gt = 341.45 m/s - 9.81 m/s^{2}*90 s = -541.45 m/s [/tex]
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!