An attacker at the base of a castle wall 4.00 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 4.00 m high throws a rock straight up with speed 5.50 m/s from a height of 1.65 m above the ground.

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

Yes No     

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

Your response differs from the correct answer by more than 10%. Double check your calculations. 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 5.50 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 No     

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

Explanation / Answer

a) maximum height reached by the rock, Hmax = h + voy^2/(2*g)

= 1.65 + 5.5^2/(2*9.8)

= 3.18 < 4m

so, No

b) let voy is the required initial speed.

Hmax = h + voy^2/(2*g)

4 = 1.65 + voy^2/(2*9.8)

==> voy = sqrt(2*9.8*(4-1.65)

= 6.79 m/s

c) let v1 = 5.5 m/s

Apply, v2^2 - v1^2 = 2*g*(H-h)

v2 = sqrt(v1^2 + 2*g*(H-h))

= sqrt(5.5^2 + 2*9.8*(4-1.65))

= 8.74 m/s

d) yes. change in speed is same.

e) because g is same for both rocks.

but as the rock goes up it speed decreses.

when rock comes down it speed increases.

but when we consider change in speed remains same.

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