An attacker at the base of a castle wall 3.75 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.75 m high throws a rock straight up with speed 8.50 m/s from a height of 1.60 m above the ground.

(a) what is the rock speed at the top? ?

m/s

(b) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 8.50 m/s and moving between the same two points.

?m/s

(c) 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? Explain physically why it does or does not agree.

Explanation / Answer

b)

Vf2=Vi2+2a(Yf-Yi)

Vf2 =8.52+2*(-9.8)*(3.75-1.6)

Vf2=30.11

Vf=5.5 m/s

c)

Vf2=Vi2+2a(Yf-Yi)

Vf2 =8.52+2*(-9.8)*(1.6-3.75)

Vf2=114.39

Vf=-10.7 m/s

Minus sign indicate motion is downward

change in speed

dV=10.7-8.5

dV=2.2 m/s

d)

Change in speed

dV=5.5-8.5 =-3 m/s

so does not agree with C

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.