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A ball is thrown straight up with an initial speed of 40 ms How high does it go?
Wiki User
∙ 9y ago
The ball reaches a maximum height of ~81.6 meters.
How to do it:
Find the change in time using Vf = Vi -g(t)
We know: Vf = 0 (terminal velocity) and Vi = 40 (given in question)
So, Vf = Vi - g(t) => 0 = 40 - g(t) --Rework the equation to solve for t...
t = 40/g => since g = 9.8... t = 40/9.8 ~ 4.08
So the change in time is about 4.08 seconds
Now plug your new values into the kinematics equation
Yf = Yi + Vi(t) + 1/2(a)(t^2)
Plug in our values and solve => Yf = 0 + (40 * 4.08) - (9.8 * 4.08^2)/2
Therefore, Yf = 81.6 meters. This is your maximum height. Remember to find the time taken for the ball to fall from this height back to the ground if you need to calculate the total time of the fall.
Source: I recently took a test with this exact question, got it right.
Thanks for posting!
~Secret Physics Man
Wiki User
∙ 9y ago
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