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iren [92.7K]
3 years ago
12

Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then

Physics
1 answer:
Andreas93 [3]3 years ago
5 0

Answer:

The displacement is  \Delta H =    -  1 \ m

The distance  is  D =  4 \  m

Explanation:

From  the question we are told that

    The height from which the ball is dropped is  h  =  1 \ m

    The height attained at  the first bounce is  h_1  = 0.8  \  m

    The height attained at  the second bounce is   h_2 = 0.5 \  m

    The height attained at  the third bounce is h_3 = 0.2 \  m

Note  : When calculating displacement we consider the direction of motion

Generally given that upward is positive  the total displacement of the ball is mathematically represented as

            \Delta H =  (0  -  h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)

Here the 0 show that there was no bounce back to the point where Billy released the ball  

           \Delta H =  (0  -  1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)

=>          \Delta H =    -  1 \ m

Generally the distance covered by the ball is mathematically represented as  

                D =  h +  2h_2 + 2h_3 + 2h_3

The 2 shows that the ball traveled the height two times

              D =  1 +  2* 0.8  + 2* 0.5 + 2* 0.2

=>           D =  4 \  m

     

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Two boats start together and race across a 48-km-wide lake and back. boat a goes across at 48 km/h and returns at 48 km/h. boat
jolli1 [7]

Answer:

Time required by boat 1 for the round trip is less than that of boat 2.

Hence, boat 1 wins.

Explanation:

Case 1: Boat 1

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{48}

time = 1 hour

Total time taken by boat 1 is,

Total time by boat 1 = 1 hour + 1 hour = 2 hour

Total time by boat 1 = 2 hour

Total time taken by boat 1 for the round trip is 2 hour.

Case 2: Boat 2

Speed of boat = \frac{distance of river}{time}

time = \frac{distance of river}{speed of boat}

While going to another end

time = \frac{distance of river}{speed of boat}

time = \frac{48}{24}

time = 2 hour

While going back,

time = \frac{distance of river}{speed of boat}

time = \frac{48}{72}

time = 0.66 hour

Total time taken by boat 2 is,

Total time by boat 1 = 2 hour + 0.66 hour

Total time by boat 1 = 2.66 hour

Total time taken by boat 2 for the round trip is 2.66 hour.

Time required by boat 1 for the round trip is less than that of boat 2.

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5 0
3 years ago
A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
Evgen [1.6K]

Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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