Answer:
Maximum height = 500 feet
Land after = 20 seconds
Step-by-step explanation:
Let us follow the movement of the ball as it is thrown upwards. We notice that we get to a point, where the change in the height of the ball with respect to time becomes = 0. This means that the ball is at its maximum position.
Hence, mathematically, at maximum height, [tex]dh/dt = 0[/tex]
Recall,
[tex]h = 100t - 5t^2[/tex]
[tex]dh/dt=100-10t[/tex]
From our problem, we can also deduce that at the maximum height, the velocity of the ball is = 0. i.e [tex]dh/dt = 0[/tex]
hence,
[tex]0 = 100-10t\\t =10 seconds[/tex]
The time the ball takes to reach a maximum height is 10 seconds.
The next step will be to substitute this time value into the equation for the height.
[tex]h = 100(10) -5(10)^2=500 feet[/tex]
The maximum height is 500 feet
The object will land after t X 2 seconds = 10 X 2 = 20 seconds. This is because it will have to go up and come back down. This should take twice the time.
