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nika2105 [10]
2 years ago
9

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

Physics
1 answer:
horrorfan [7]2 years ago
3 0

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:

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

V=V_{o}+gt (2)

V^{2}={V_{o}}^{2}+2gy (3)

Where:

y is the height of the ball at a given time

y_{o}=0 is the initial height of the ball

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

t is the time  

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

V is the final velocity of the ball at a given time

Now let's start with the answers:

<h2>A) Total time of the ball in the air</h2>

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:

y=0 assuming the initial and the final height is zero

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

Isolating t:

t=\frac{-2V_{o}}{g} (5)

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

Then:

t=10 s (7)

<h2>B) Velocity at 7 s</h2>

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

V=V_{o}+gt

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

Hence:

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

<h2>C) Maximum height</h2>

The height of the ball has its maximum value when V=0, 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:

V^{2}={V_{o}}^{2}+2gy

0={V_{o}}^{2}+2gy (10)

Finding y:

y=\frac{-{V_{o}}^{2}}{2g} (11)

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

Then:

y=400 ft (12) This is the ball's maximum height

<h2>D) Height at 7 s</h2>

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

y=y_{o}+V_{o}t+\frac{1}{2}gt^{2}

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

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

Finally:

y=336 ft (15)

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mass of an electronic charge = 9.10 * 10^{-31} kg

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7 0
3 years ago
A block of mass M is connected by a string and pulley to a hanging mass m. The coefficient of kinetic friction between block M a
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Explanation:

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             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

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             T - fr = M a

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we add and resolved

             mg-  μ Mg = (M + m) a

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             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

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          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

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        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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