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3 years ago

7

If a basball is project upwards from the ground level with an initial velovaity of 32 feet per second, then it's height is a fun

ction of time.
Given by s= -8t^2 + 32t. what is the maximum height reached by the ball?

1 answer:

inessss [21]3 years ago

6 0

Answer:

Maximum height reached by the ball is 32 meters.

Explanation:

It is given that,

If a baseball is project upwards from the ground level with an initial velocity of 32 feet per second, then it's height is a function of time. The equation is given as :

$s=-8t^2+32t$ ...........(1)

t is the time taken

s is the height attained as a function of time.

Maximum height achieved can be calculated as :

$\dfrac{ds}{dt}=0$

$\dfrac{d(-8t^2+32t)}{dt}=0$

-16 t + 32 = 0

t = 2 seconds

Put the value of t in equation (1) as :

$s=-8(2)^2+32(2)$

s = 32 meters

So, the maximum height reached by the ball is 32 meters. Hence, this is the required solution.

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