What is the maximum height the ball reaches? A ball is thrown vertically upward from the roof of a 64 foot tall building with a velocity of 96 ft/sec, its height in feet after t seconds is s(t) = 64...
What is the maximum height the ball reaches?
A ball is thrown vertically upward from the roof of a 64 foot tall building with a velocity of 96 ft/sec, its height in feet after t seconds is s(t) = 64 + 96 t - 16 t^2.
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The ball is thrown upwards at 96 ft/s from the roof of a building which is 64 foot tall.
The height of the ball after t seconds is given by the function:
s(t) = 64 + 96t - 16t^2
To find the maximum height of the ball, we first find the derivative of the function that defines the height.
s'(t) = 96 - 32t
Equate this to 0,
96 - 32t = 0
=> t = 96/32
=> t = 3
We also see that s''(t) = -32, which is negative, confirming that at t = 3 we have a maximum value.
At time equal to 3 seconds, the height of the ball is s(3) = 64 + 96*3 - 16*9
=> 208
The required maximum height of the ball is 208 feet.
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