Physics, 31.01.2020 13:51 dylannhandy

If a ball is thrown vertically upward with a velocity of 144 ft/s, then its height after t seconds is s = 144t − 16t2. (a) what is the maximum height reached by the ball? ft (b) what is the velocity of the ball when it is 320 ft above the ground on its way up? (consider up to be the positive direction.) ft/s what is the velocity of the ball when it is 320 ft above the ground on its way down? ft/s

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The answers for the ball thrown vertically upward with a velocity of 144 ft/s and with a height after t seconds of s = 144t - 16t² are:

a) The maximum height reached by the ball is 324 ft.

b) The velocity of the ball when it is 320 ft above the ground on its way up is 16 ft/s.

c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s.

a) The maximum height reached by the ball can be calculated with the given equation:

$s = 144t - 16t^{2}$ (1)

Where:

s: is the height

t: is the time

We can find the time with the following equation:

$v_{f} = v_{i} - gt$ (2)

Where:

$v_{f}$ : is the final velocity = 0 (at the maximum height)

$v_{i}$ : is the initial velocity = 144 ft/s

g: is the acceleration due to gravity = 32 ft/s²

Solving equation (2) for t and entering into equation (1), we can find the maximum height:

$s = 144t - 16t^{2} = 144(\frac{v_{i}}{g}) - 16(\frac{v_{i}}{g})^{2} = 144(\frac{144 ft/s}{32 ft/s^{2}}) - 16(\frac{144 ft/s}{32 ft/s{2}})^{2} = 324 ft$

Hence, the maximum height is 324 ft.

b) To find the velocity of the ball when it is 320 ft above, we can use the following equation:

$v_{f}^{2} = v_{i}^{2} - 2gs$

$v_{f}^{2} = (144 ft/s)^{2} - 2*32 ft/s^{2}*320 ft$

The above equation has two solutions:

$v_{f_{1}} = 16 ft/s$

$v_{f_{2}} = -16 ft/s$

Since the question is for the velocity of the ball on its way up and considering the way up as the positive direction, the answer is the positive value $v_{f_{1}} = 16 ft/s$ .

c) The velocity of the ball when it is 320 ft above the ground on its way down is -16 ft/s (we take the negative value calculated above, $v_{f_{2}}$ ). We consider the way down as the negative direction.

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I hope it helps you!

Answer from: jacobd578

(a) 168.2 ft/s

The vertical position of the ball is given by

s = 144t - 16t^2

where t is the time.

By differentiating this expression, we find the velocity:

v = 144-32 t

The maximum height is reached when the velocity is zero, so:

0 = 144 - 32 t

From which we find

$t = \frac{144}{32}=4.5 s$

And substituting this value into the equation for s, we find the maximum height:

s = 144(4.5 s)-16(4.5 s)^2=324 ft

(b) 16 ft/s

We want to find the velocity of the ball when the position of the ball is

s = +320 ft

Substituting into the equation for the position,

320 = 144t-16t^2

16t^2 -144t +320 = 0

Solving for t, we find two solutions:

t = 4 s

t = 5 s

The first one corresponds to the instant in which the ball is still on its way up: Substituting into the equation for the velocity, we find the velocity of the ball at that time

v = 144 - 32 t=144- 32(4 s)=16 ft/s

Now we want to find the velocity of the ball when the position of the ball is

s = +320 ft

but on its way down. In the previous part, we found

t = 4 s

t = 5 s

So the second time corresponds to the instant in which the ball is at s = 320 ft but on the way down.

Substituting t = 5 s into the equation for the velocity, we find:

v = 144 - 32 t=144- 32(5 s)=-16 ft/s

And the negative sign means the direction is downward.

If a ball is thrown vertically upward with a velocity of 144 ft/s, then its height after t seconds i

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