Physics, 14.09.2019 01:30 Kaesy24

Astudent at the top of a building of height h throws ball a straight upward with speed v0 (3 m/s) and throws ball b straight downward with the same initial speed. a. compare the balls’ accelerations, both direction, and magnitude, immediately after they leave her hand. is one acceleration larger than the other? or are the magnitudes equal? b. compare the final speeds of the balls as they reach the ground. is one larger than the other? or are they equal?

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The magnitude of the acceleration of both balls immediately after they left her hand is the same but the direction is different.

The velocity of the ball thrown upwards will be larger than the velocity of the ball thrown downwards when they reach the ground.

The given parameters;

height of the building = hinitial velocity of the ball A = v₀initial velocity of the ball B, = v₀

Both balls experiences acceleration due to gravity, given in the following equation;

$h_{up} = v_ot - \frac{1}{2} gt^2\\\\h_{down} = v_0t + \frac{1}{2} gt^2$

Thus, the magnitude of the acceleration of both balls immediately after they left her hand is the same but the direction is different.

The final velocity of the ball dropped downwards is calculated as;

$v_f^2 = v_0 + 2gh\\\\v_f = \sqrt{v_0 + 2gh}$

The maximum height reached by the ball projected upwards is calculated as follows;

$h_o = v_0 - \frac{1}{2} gt^2$

The final velocity of the ball thrown upwards before it hits the ground;

$v_f^2 = v_0 ^2 + 2gh\\\\v_f^2 = 0 + 2g(h_0 + h)\\\\v_f^2 = 2gh + 2gh_0\\\\v_f^2 = 2gh + 2g(v_0 - \frac{1}{2} gt^2)\\\\v_f^2 = 2gh + 2gv_0 - g^2t^2\\\\v_f = \sqrt{ 2gh + 2gv_0 - g^2t^2$

Thus, the velocity of the ball thrown upwards will be larger than the velocity of the ball thrown downwards when they reach the ground. The direction of both velocities is the same when they reach the ground.

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Answer from: nina4552

same

Explanation:

Given

ball A initial velocity=3 m/s(upward)

Ball B initial velocity=3 m/s (downward)

Acceleration on both the balls will be acceleration due to gravity which will be downward in direction

Both acceleration is equal

For ball A

maximum height reached is $h_1=\frac{3^2}{2g}$

After that it starts to move downwards

thus ball have to travel a distance of h_1+h(building height)

so ball A final velocity when it reaches the ground is

$v_a^2=2g\left ( h_1+h\right )$

$v_a^2=2g\left ( 0.458+h\right )$

$v_a=\sqrt{2g\left ( 0.458+h\right )}$

For ball b

$v_b^2-\left ( 3\right )^2=2g\left ( h\right )$

$v_b^2=2g\left ( \frac{3^2}{2g}+h\right )$

$v_b=\sqrt{2g\left ( 0.458+h\right )}$

thus v_a=v_b

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Astudent at the top of a building of height h throws ball a straight upward with speed v0 (3 m/s) an...

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