Painevg

2022-01-10

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is $s=80t-16{t}^{2}$. What is the velocity of the ball when it is 96 ft above the ground on its way up? On its way down?

Step 1
We first need to find the times at which the height of the ball is 96ft
h(t)=96
$80t-16{t}^{2}=96$
Subtract 96 from both sides and rewrite in the standard from:
$-16{t}^{2}=80t-96=0$
Divide both sides by -16
${t}^{2}-5t+6=0$
Factorize
(t-2)(t-3)=0
Using the zero-product property, we can write t-2=0, t=2
t-3=0, t=3
Step 2
Since the rock goes up first and then it comes down, the smaller time value corresponds to the velosity of the rock going up.
By differentiating h(t)using the power rule, we get the velocity for the ball
v(t)=80-32t
Substitute t=2, to get the velocity of the rocket at a height of 96ft while it is going up
$v\left(2\right)=80-32\cdot 2=16\frac{ft}{s}$
Substitute t=3, to get the velosity of the rocket at a height of 96ft while it is coming down
$v\left(3\right)=80-32\cdot 3=-16ft/s$
Result
The velosity when going up is 19 ft/s, while comig down it is -16 ft/s

Mary Nicholson

Step 1
Set s=96 and find the times at which the projectile has a height of 96. t=2 represents the time at which the projectile is moving upwards, and t=3 downwards.
$s=80t-16{t}^{2}$
$96=80t-16{t}^{2}$
$0={t}^{2}-5t+6$
t=2,3
Step 2
Take the derivative of position to find velocity as a function of time. Find v(2) and v(3)
$v\left(t\right)=\frac{ds}{dt}=80-32t$
$v\left(2\right)=80-32\left(2\right)=16\frac{ft}{s}$
$v\left(3\right)=80-32\left(3\right)=-16\frac{ft}{s}$
Result
$16\frac{ft}{s}$
$-16\frac{ft}{s}$

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