## What is the initial value of h(t)= -9.8t2+48t+50?

Question

What is the initial value of h(t)= -9.8t2+48t+50?

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6 months 2021-08-09T14:35:43+00:00 1 Answers 4 views 0

The height h of an object after t seconds is

h=-16t^2+48t+210h=−16t

2

+48t+210

The height of a neighboring 50-foot tall building is modeled by the equation h=50.

The time (t) when the object will be at the same height as the building is found to be t = –2 and t = 5.

To find:

The statement which describes the validity of these solutions.

Solution:

We have,

h=-16t^2+48t+210h=−16t

2

+48t+210

Here, t is the time in seconds.

For t=-2,

h=-16(2)^2+48(-2)+210h=−16(2)

2

+48(−2)+210

h=-64-96+210h=−64−96+210

h=50h=50

For t=5,

h=-16(5)^2+48(5)+210h=−16(5)

2

+48(5)+210

h=-400+240+210h=−400+240+210

h=50h=50

So, the value of h is 50 at t=-2 and t=5.

We know that time is always positive so it cannot be negative value. It means t=-2 is not possible.

The solution t = 5 is the only valid solution to this system since time cannot be negative.

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