An attacker at the base of a castle wan 3.55 m high throws a rock straight up wi

Question

An attacker at the base of a castle wan 3.55 m high throws a rock straight up with speed 4.50 m/s from a height of 1.70 m above the ground. Will the rock reach the top of the wall? Yes no If so, what is its speed at the top? If not, what initial speed must It have to reach the top? Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 4.50 m/s and moving between the same two points. Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Yes No Explain physically why it does or does not agree.

Explanation / Answer

a)

maximum height reached, h = u^2/2g

= 4.5^2/(2*9.8)

= 1.03 m

So, maximum height reached from ground = 1.03+1.7 = 2.73 m (which is < 3.55 m)

So, the rock ont reach the top of the wall.

b)

For reaching the top, the height it mst go up = 3.55 - 1.7 = 1.85 m

So, the velocity needed, u = sqrt(2gh)

= sqrt(2*9.8*1.85)

= 6.02 m/s

c)

v = sqr( u^2 + 2gh)

= sqrt(4.5^2 + 2*9.8*1.85)

= 7.52 m/s

So, change in speed = 7.52 - 4.5 = 3.12 m/s

d)

No, change of speed does not agree.

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