An object travels through the air following a

Shahreen Perveen

Shahreen Perveen

Answered question

2022-09-27

An object travels through the air following a path By ht=-4.9t2+17t, where h is the height in meter, above the ground and t is the time. How long to the nearest tenth of a secon, is the object in the air?

Answer & Explanation

Nick Camelot

Nick Camelot

Skilled2023-06-05Added 164 answers

To find how long the object is in the air, we need to determine the time when the height h(t) is equal to zero. We can set h(t)=0 and solve for t.
The equation representing the height of the object above the ground is given by:
h(t)=4.9t2+17t
Setting h(t) = 0, we have:
4.9t2+17t=0
To solve this quadratic equation, we can factor out common terms:
t(4.9t+17)=0
This equation is satisfied when either t = 0 or 4.9t+17=0.
For t=0, the object is at the ground level. However, we are interested in the time when the object is in the air, so we focus on solving 4.9t+17=0:
4.9t+17=0
Adding 4.9t to both sides:
17=4.9t
Dividing both sides by 4.9:
174.9=t
Calculating the value:
t3.469
Therefore, the object is in the air for approximately 3.469 seconds (rounded to the nearest tenth of a second).

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Math Word Problem

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?