Question 311209: A water ball catapult is plased on top of schoolbuilding 25m above soccer field a launched balloon height is -5(t^2)+20t+25 What time will it hit the ground and what is the greatest height it went up to before hitting the ground?
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! To find the time when it will hit the ground, set your function to 0 and solve for t.
-5(t^2)+ 20t + 25 = 0
Can you solve for t?
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I read your question again. They gave you the position equation, and increased the gravitational constant to make the math easier.
For time, solve the position equation y = -5(t^2)+20t+25 where y = 0 (ground level). You'll get two solutions. Discard the negative one.
For the maximum hight, you want to know when the ball just starts coming down, ie, when it's vertical velocity is zero. The velocity equation is the first derivative in the position equation. Solve that where dy/dt = 0 to get the time at maximum height, and plug that time back into the position equation.
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