An attacker at the base of a castle wall 3.85 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.85 m high throws a rock straight up with speed 4.00 m/s from a height of 1.45 m above the ground.

(a) Will the rock reach the top of the wall?

Yes or No

(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?
m/s

(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 4.00 m/s and moving between the same two points.
m/s

(d) 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 or No    

(e) Explain physically why it does or does not agree.

Explanation / Answer

a) No

b) since we know that

Maximum height = u^2/2g

3.85-1.45 = u^2/2g

47.04 = u^2

u = 6.858 m/sec

c) conserving energy

0.5mu^2 +mg (1.45 ) = 0.5mV^2

V = SQRT ( 4^2 +2*1.45*9.8)

V = 6.664 m/sec

change in speed

V-u = 2.66 m/sec

d) No

e ) from conservation of energy for same decrease in potential energy there will be same increase in kinetic energy

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