Question

An object is launched vertically in the air at 24.5 meters per second from a 11-meter-tall platform. Using the projectile motion model h(t)=-4.9t^2+v0t+h0, where h(t) is the height of the projectile t seconds after it’s departure, v0 is the initial velocity in meters per second, and h0 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:

2.5 is a time to reach at maximum height and 41.625 m is maximum height of object.

Step-by-step explanation:

$\text{Given model:}h(t)=-4.9t^2+v_0t+h_0$

Where,

$v_0$ is initial velocity = 24.5 m/s

$h_0$ is initial height = 11 m

h(t) is the height of the projectile after t seconds.

We need to find the maximum height and time to reach maximum height.

$h(t)=-4.9t^2+24.5t+11$

It is parabolic equation. So, maximum height at vertex. Now we calculate the vertex of h(t)

$t=-\frac{b}{2a}$

where, a=-4.9 and b=24.5

$t=-\frac{24.5}{-4.9\times 2}$

t=2.5 seconds

It will take 2.5 seconds to reach maximum height.

Now we put t=2.5 s into h(t) to find maximum height.

$h(2.5)=-4.9(2.5)^2+24.5(2.5)+11=$

Maximum height is 41.625 m

Thus, 2.5 seconds is a time to reach at maximum height and 41.625 m is maximum height of object.

Answer 2

The maximum height is 41.625 meters and it occurs at 2.5 seconds.

Given that the height of an object (h) at time t is given by:

h(t)=-4.9t² +vt + h

Initial velocity = 24.5 m/s, initial height (h) =  11 m, hence:

h(t)=-4.9t² +24.5t + 11

The maximum height is at dh/dt = 0, hence:

dh/dt = -9.8t + 24.5

-9.8t + 24.5 = 0

9.8t = 24.5

t = 2.5 seconds

h(2.5) = -4.9(2.5)² +24.5(2.5) + 11 = 41.625 m

Hence the maximum height is 41.625 meters and it occurs at 2.5 seconds.

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