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d1i1m1o1n [39]
3 years ago
8

While running these tests, crall notices a similarity in the velocity measured at the ground independent of whether the tennis b

all is thrown upward or downward. she decides to see if her hypothesis that the final speed is independent of the sign of the initial velocity is correct. crall and whipple decide to run the experiment several times using different initial speeds of the elevator each time comparing the final velocity of the tennis ball when the elevator is traveling upward vs. downward at the same speed?
Physics
2 answers:
lara [203]3 years ago
6 0

At the ground the ball will always have velocity along the direction of gravity. If upward motion is taken positive it will always have negative velocity at the ground because, if the ball was given an initial upward velocity then gravity will decelerate it and bring it down with a negative final velocity. If the ball is given an initial downward velocity then the ball will be further accelerated by gravity in the downward direction only, again maintaining negative direction. The magnitude however in both cases will be different. the final velocity at the ground will have higher magnitude in case of elevator moving downwards.

lora16 [44]3 years ago
6 0

Answer:

The velocities will be different

Explanation:

Thinking process:

The downward motion is more or less the same. This is because the gravitational pull on the object. In a vacuum, the terminal speed of an object is the same throughout.

Throwing the ball downward simply converts the potential energy to kinetic energy. However, throwing the ball up needs more energy to overcome the gravitational pull on the ball. Hence the velocity of the ball will be different.

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Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

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