1) A freely falling object requires 1.90 s totravel the last 22.0 m before it hi

Question

1)
A freely falling object requires 1.90 s totravel the last 22.0 m before it hits theground. From what height above the ground did it fall?
2)
An attacker at the base of a castle wall 3.97 m high throws a rock straight up with speed6.00 m/s at a height of 1.45 m above the ground.
a)If the rock will reach the top, what is its speed? If it willnot, what initial speed must it have to reach the top?
b)Find the change in speed of a rock thrown straight down from thetop of the wall at an initial speed of 6.00 m/s and moving between the same two points.

Explanation / Answer

Take the upward direction as positive. the acceleration is a = -g = -9.8 m/s2 Take the origin as point A, the height of 22.0 m as point B,and the ground as point C. If we assume point B is the starting point, where the velocityis v. then the height of the object is h = vt + 0.5at2 + h0 = vt-4.9t2 + 22 we know that it takes 1.9 s to hit the ground, so t =1.9, h = 0, 0 = 1.9v - 4.9 * 1.92 + 22 v = -15.099 m/s Then when the object hits the ground, its velocity is vf = v + at = -15.099 + (-9.8) * 1.9 = -33.719m/s. Now go back to the origin of the free fall. using conservationof energy mgh = 0.5m(vf )2 9.8 h = 0.5 * (-33.719)2 h = 58.008 meters. so the object falls from 58.008 meters high. h = 58.008 meters. so the object falls from 58.008 meters high.

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