Shahreen Perveen

2022-09-27

An object travels through the air following a path By $h\left(t\right)=-4.9{t}^{2}+{17}^{t}$, 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?

Nick Camelot

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 $h\left(t\right)=0$ and solve for t.
The equation representing the height of the object above the ground is given by:
$h\left(t\right)=-4.9{t}^{2}+17t$
Setting h(t) = 0, we have:
$-4.9{t}^{2}+17t=0$
To solve this quadratic equation, we can factor out common terms:
$t\left(-4.9t+17\right)=0$
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 $-4.9t+17=0$:
$-4.9t+17=0$
$17=4.9t$
Dividing both sides by 4.9:
$\frac{17}{4.9}=t$
Calculating the value:
$t\approx 3.469$
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?