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A ball is thrown straight up in the air at a velocity
of 100 feet per second. The height of the ball at t seconds
is given by the formula: h = 100t – 5t^2 Find the
maximum height of the ball and when it will land. Maximum Height ____ ft. Land after: ___ seconds


Sagot :

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.