Question

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?

Answer 1

Answer:

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

$h(t)=-4.9t^2+v_{0}t+h_{0}$

Put the value in the equation

$h(t)=-4.9t^2+41.65t+99$

On differentiating equation (I)

$\dfrac{dh}{dt}=v=-9.8t+41.65$

We need to calculate the time at maximum height

The velocity is zero at maximum height,

So, $-9.8t+41.65=0$

$t =\dfrac{-41.65}{-9.8}$

$t=4.25\ sec$

We need to calculate the maximum height

Put the value of t in equation (I)

$h_(max)=-4.9\times(4.25)^2+41.65\times4.25+99$

$h_{max}=187.50\ m$

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

The maximum height is 187.50 m.