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A ball is given an initial velocity of 160ft/s straight up. Use g = -32ft/s^2

A. Find the total time the ball is in the air
B. Find the velocity of the ball in ft/s at 7s
C. Find the maximum height of the ball in ft
D. Find the height of the ball above ground in feet at 7s

Respuesta :

cryssatemp

Answers:

A) 10 s

B) -64 ft/s

C) 400 ft

D) 336 ft

Explanation:

This described situation is related to vertical motion, and the main equations for this situation are as follows:

[tex]y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}[/tex] (1)

[tex]V=V_{o}+gt[/tex] (2)

[tex]V^{2}={V_{o}}^{2}+2gy[/tex] (3)

Where:

[tex]y[/tex] is the height of the ball at a given time

[tex]y_{o}=0[/tex] is the initial height of the ball

[tex]V_{o}=160 ft/s[/tex] is the initial velocity of the ball

[tex]t[/tex] is the time  

[tex]g=-32 ft/s^{2}[/tex] is the acceleration due to gravity on Earth (directed downwards)

[tex]V[/tex] is the final velocity of the ball at a given time

Now let's start with the answers:

A) Total time of the ball in the air

In this case we will use equation (1) to calculate the total time the ball was in the air (since it was thrown straight up until it hit the ground) with the following condition:

[tex]y=0[/tex] assuming the initial and the final height is zero

[tex]0=0+V_{o}t+\frac{1}{2}gt^{2}[/tex] (4)

Isolating [tex]t[/tex]:

[tex]t=\frac{-2V_{o}}{g}[/tex] (5)

[tex]t=\frac{-2(160 ft/s)}{-32 ft/s^{2}}[/tex] (6)

Then:

[tex]t=10 s[/tex] (7)

B) Velocity at 7 s

In this part we will use equation (2) in order to find the final velocity of the ball when [tex]t= 7 s[/tex]:

[tex]V=V_{o}+gt[/tex]

[tex]V=160 ft/s+(-32 ft/s^{2})(7 s)[/tex] (8)

Hence:

[tex]V=-64 ft/s[/tex] (9)  The negative sign indicates the velocity is directed downwards

C) Maximum height

The height of the ball has its maximum value when [tex]V=0[/tex], just in the moment at the top of its movement, before the begining of the free fall.

In this case we will use equation (3) with the explained condition above:

[tex]V^{2}={V_{o}}^{2}+2gy[/tex]

[tex]0={V_{o}}^{2}+2gy[/tex] (10)

Finding [tex]y[/tex]:

[tex]y=\frac{-{V_{o}}^{2}}{2g}[/tex] (11)

[tex]y=\frac{-(160 ft/s)^{2}}{2(-32 ft/s^{2})}[/tex] (12)

Then:

[tex]y=400 ft[/tex] (12) This is the ball's maximum height

D) Height at 7 s

In this part we can use equation (1) for [tex]t=7 s[/tex]:

[tex]y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}[/tex]

[tex]y=0+V_{o}t+\frac{1}{2}gt^{2}[/tex] (13)

[tex]y=(160 ft/s)(7s)+\frac{1}{2}(-32 ft/s^{2})(7s)^{2}[/tex] (14)

Finally:

[tex]y=336 ft[/tex] (15)

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