SOLUTION: A ball is thrown straight upward at an initial speed of 149 feet per second.
h=-16⁢t2+149⁢t
When does the ball reach the highest point?
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h=-16⁢t2+149⁢t
When does the ball reach the highest point?
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Question 339928: A ball is thrown straight upward at an initial speed of 149 feet per second.
h=-16t2+149t
When does the ball reach the highest point? Found 2 solutions by nerdybill, solver91311:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! Given:
h = -16t^2 + 149t
.
The time (t) when it reaches the "ighest" point would be at the "axis of symmetry":
t = -b/2a = -149/(2*(-16)) = -149/(-32) = 4.66 sec
is only true if the ball was launched from the ground at 149 feet per second initial velocity. Typically when you throw something upward, the ball is above the ground by a distance that is in proportion to the thrower's height.
Be that as it may, the function describes a parabola opening downward. The maximum height is reached at the value of for the vertex of the parabola. Use where and
By the way, 149 feet per second is roughly 101.5 miles per hour, very close to the top speed of a pitched baseball. Find me the guy who can throw a ball straight up at that velocity.
John
My calculator said it, I believe it, that settles it
