Rachel's crew is finally successfully in shooting the firework. They shoot it to its maximum height in 7 seconds and it explodes sending fragments scattering in horizontal directions all around at a speed of 20 meters per second. In answering the questions below,assume that air drag is negligible. What will be the acceleration of the fragments as they fall vertically? Assuming that they all fall from the same maximum height,how long will it take the fragments to hit the ground?
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1>Acceleration will be due to gravity = 9.8 m/sec^2
2>>As the fragment reach a maximum height in 7 sec.So as they are falling from same height then they will take the same time to reach down also.
1>Acceleration will be due to gravity = 9.8 m/sec^2
2>>As the fragment reach a maximum height in 7 sec.So as they are falling from same height then they will take the same time to reach down also.
nita456:
Can you explain the second answer more?
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velocity of mass at the maximum height = 0 m/s
height = h = 7 m.
the fragments are projected in horizontal direction with velocity = 20 m/s
In the horizontal direction, there is no acceleration. In the vertical direction the acceleration = g = acceleration due to gravity.
since the initial vertical velocity is 0 m/s,
h = 0 t + 1/2 g t²
t = √(2h/g) = time for the fragments to reach the ground.
in this time, the fragments travel 20 * √(2h/g) meters horizontally.
height = h = 7 m.
the fragments are projected in horizontal direction with velocity = 20 m/s
In the horizontal direction, there is no acceleration. In the vertical direction the acceleration = g = acceleration due to gravity.
since the initial vertical velocity is 0 m/s,
h = 0 t + 1/2 g t²
t = √(2h/g) = time for the fragments to reach the ground.
in this time, the fragments travel 20 * √(2h/g) meters horizontally.
the acceleration will be equal to g = 9,81 m/sec^2 . this is the acceleration due to gravity.
we have the equation : v^2 - u^2 = 2 g h , here u = initial velocity. and v = final velocity. h vertical distance traveled.
when the object travels upwards, at the maximum height, v = 0. so height reached is h = v^2 / 2g. when the object travels downwards, u= 0. so final velocity at the ground = v^2 = 2 g h. thus the initial velocity of projection of object, is the same as the final velocity of the object when it comes back to ground from the top position it attains.
how, similarly, v = u + g t . on the way up, v = 0. so we have t = v/g.
on the way down we have u = 0, so v = gt => t = v/g and the value u for upward journey is same as the v for the downward journey. Hence, the times of flight upwards and downwards are equal.
So could you explain the accelerati
So they acceleration would be 9.8 m/s2
So how long will it take the fragments to hit the ground and how did you come up with that answer?
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