An object is launched vertically in the air at 41.6541.65 meters per second from a 99-meter-tall platform. using the projectile motion model h (t )equals negative 4.9 t squared plus v 0 t plus h 0 commah(t)=−4.9t2+v0t+h0, where h(t) is the height of the projectile t seconds after its departure, v 0v0 is the initial velocity in meters per second, and h 0h0 is the initial height in meters, determine how long it will take for the object to reach its maximum height. what is the maximum height?

The time when the object to reach its maximum height is 4.25 sec.

The maximum height is 187.50 m.

Explanation:

Given that,

Initial velocity = 41.65 m/s

Height = 99 m

The projectile motion model h(t) is

Put the value in the equation

On differentiating equation (I)

We need to calculate the time at maximum height

The velocity is zero at maximum height,

So,

We need to calculate the maximum height

Put the value of t in equation (I)

Hence, The time when the object to reach its maximum height is 4.25 sec.

The maximum height is 187.50 m.

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