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valentinak56 [21]
2 years ago
6

A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5m long. The bowler hears the sound o

f the ball hitting the pins 2.50s after the ball is released from his hands. What is the speed of the ball? The speed of the sound is 340 m/s.
Physics
1 answer:
kobusy [5.1K]2 years ago
6 0
  <span> The ball takes 15/V secs to reach the pin since T=D/V. so there for the  Sound takes 15/340 secs to reach the bowler. The total time enlapsed then is 15/V + 15/340 = 2.10 So 15/V =2.1 -15/340
V = 7.296 m/s </span>
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A bullet is fired with a velocity of 100 m/s from the ground at an angle of 60° with the horizontal. Calculate the horizontal ra
seraphim [82]

1) The horizontal range of the bullet is 884 m

2) The maximum height attained by the bullet is 383 m

Explanation:

1)

The motion of the bullet is a projectile motion, which consists of two separate motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

From the equation of motions along the x- and y- directions, it is possible to find an expression for the horizontal range covered by a projectile, and it is:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial speed of the projectile

\theta is the angle of projection

g=9.8 m/s^2 is the acceleration of gravity

For the bullet in the problem, we have

u = 100 m/s (initial speed)

\theta=60^{\circ} (angle)

Solving the equation, we find the horizontal range:

d=\frac{(100)^2sin(2\cdot 60^{\circ})}{9.8}=884 m

2)

To find the maximum height, we have to analyze the vertical motion of the bullet. We can do it by using the following suvat equation:

v_y^2 - u_y^2 = 2as

where

v_y is the vertical velocity of the bullet after having covered a vertical displacement of s

u_y is the initial vertical velocity

a =-g= is the acceleration (negative, since it points downward)

The vertical component of the initial velocity is given by

u_y = u sin\theta

Also, the maximum height s is reached when the vertical velocity becomes zero,

v_y =0

Substituting into the equation and re-arranging for s, we find the maximum height:

s=\frac{u^2 sin^2 \theta}{2g}=\frac{(100)^2(sin 60^{\circ})^2}{2(9.8)}=383 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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3 years ago
A 66-kg fellow stands on a digital scale in an elevator that accelerates upwards from rest to 4.5 m/s in 2.00 s. show answer Inc
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Answer:

Explanation:

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a = 4.5 /2

= 2.25 m / s²

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Applying Newton's law of motion

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Explanation:

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