Respuesta :
Answer: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 [tex]h_1=\frac{3^2}{2g}[/tex]
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
[tex]v_a^2=2g\left ( h_1+h\right )[/tex]
[tex]v_a^2=2g\left ( 0.458+h\right )[/tex]
[tex]v_a=\sqrt{2g\left ( 0.458+h\right )}[/tex]
For ball b
[tex]v_b^2-\left ( 3\right )^2=2g\left ( h\right )[/tex]
[tex]v_b^2=2g\left ( \frac{3^2}{2g}+h\right )[/tex]
[tex]v_b=\sqrt{2g\left ( 0.458+h\right )}[/tex]
thus [tex]v_a=v_b[/tex]
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 = h
- initial velocity of the ball A = v₀
- initial velocity of the ball B, = v₀
Both balls experiences acceleration due to gravity, given in the following equation;
[tex]h_{up} = v_ot - \frac{1}{2} gt^2\\\\h_{down} = v_0t + \frac{1}{2} gt^2[/tex]
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;
[tex]v_f^2 = v_0 + 2gh\\\\v_f = \sqrt{v_0 + 2gh}[/tex]
The maximum height reached by the ball projected upwards is calculated as follows;
[tex]h_o = v_0 - \frac{1}{2} gt^2[/tex]
The final velocity of the ball thrown upwards before it hits the ground;
[tex]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[/tex]
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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