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deff fn [24]
3 years ago
10

A 0.0121-kg bullet is fired straight up at a falling wooden block that has a mass of 4.99 kg. The bullet has a speed of 898 m/s

when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.
Physics
1 answer:
aliya0001 [1]3 years ago
7 0

Answer:

t = 0.11 s

Explanation:

The motion is complete inelastic so the motion can be model using

mb * vb + mB * vB = ( mb + mB ) vf

mb = 0.0121 kg ,  mB = 4.99 kg ,  vb = 898 m / s , vB = 0

Replacing to find vf

0.0121 * 898 m / s + 4.99 kg * 0 = ( 0.0121 + 4.99) * vf

vf = 2.18 m / s

Now to find the time take the acceleration as a= g = 9.8 m /s²

t = mb * vb / a * (mb + 2 * mB)

t = [0.0121 kg * 898 m /s] / 9.8 * ( 0.0121 kg + 2*4.99)

t = 0.11 s

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14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball
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The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

  • Height h = 75 m
  • Initial velocity u = 0 ( vertical velocity )
  • Acceleration due to gravity g = 9.8 m/s²
  • Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

Learn more about Vertical motion here: brainly.com/question/24230984

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