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soldi70 [24.7K]
3 years ago
8

Jacob is traveling at 5.00 m/s North. Jacob throws a ball with a velocity of 5.00 m/s South. Jacob throws the ball from a height

of 1.45 m above the ground. Assuming gravitational acceleration is 9.81 m/s^2, how long does the ball stay in the air after it is thrown? Describe the motion of the ball relative to the ground. Describe the motion of the ball relative to Jacob.
Physics
1 answer:
Nataly [62]3 years ago
3 0

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

s=ut+\frac{gt^2}{2}

1.45=0+\frac{9.81\times t^2}{2}

t^2=frac{2\times 1.45}{9.81}

t=0.54 s

Motion of ball will be straight line

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a)  v = √(v₀² + 2g h),    b)      Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

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in this case the height y is zero and the height i = h

        v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

         v = √(v₀² + 2g h)

The times to get to the ground

ball 1

         v = v₀ - g t₁

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ball 2

         v =  -v₀ - g t₂

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From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

       Δt = t₂ -t₁

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6 0
3 years ago
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Answer:

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159kg × 4.9

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