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?
- 9 years ago
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.
- SueLv 79 years ago
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=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.
- Anonymous9 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 groundSource(s): my peter parker