# A ball is thrown upward with an upward?

A ball is thrown upward with an upward with an inital velocity of 25 feet per second. The height of the ball is presented by the equation h(t) = 25t - 16t^2, where t represents the time and h(t) represents the height. When would the ball hit the ground?

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• The ball would hit the ground when the height is zero, h(t) = 0

25t - 16t^2 = 0

t( 25 - 16t) = 0

There will be 2 values of t

t = 0 , t = 16/25seconds

t= 0 is ignored as that value of t denotes the initial time when the ball was still, in the original position

therefore, the ball, after it is thrown upwards, it will go back to it's original position (ground) at 16/25 seconds. That is the time where the ball will hit the ground.

• The ground is h(t) = 0. So this problem is really asking you to set P(x) = 0 and solve for x.

25t -16t² = 0

t(25t -16)=0

t=0 is extraneous (an unused answer)

25t - 16= 0

25t = 16

t= 16/25 sec. after the ball is released it will hit the ground.

• Anonymous
9 years ago

as you might know that, (v = u + at)

,final velocity (at a point above the ground will be 0)=0

initial velocity =25ft per sec

acceleration = -32ft per sec^2 (approx)

therefore, 0= 25+(-32) x t,

25/32sec=time upward, 25/32 sec time downward, so, total time taken is 50/32 sec for ball to reach the ground

Source(s): my peter parker