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.

Answer 1

a) Use the kinematic formula
Vf^2 = Vi^2 + 2*a*y
where Vf = 0 -- it comes to a stop at the highest point
Vi = 8.50 m/s
a = -g = -9.8 m/s^2 -- negative because the 8.50 m/s is up and gravity acts downward
y = the height at which Vf is zero. Is that high enough?

b) Does it go as high or higher than the top?
If yes, use the same formula as in part a, this time Vf is the unknown and y = 3.75 m - 1.6 m, everthing else is the same as in part a.
If no, use the same formula as in part a, this time Vi is the unknown and y = 3.75 m - 1.6 m, everthing else is the same as in part a.

c) Use the same formula as in part a, this time
Vf is the unknown
Vi = 8.50 m/s
a = g = +9.8 m/s^2 -- a is +g because the initial speed is also down
y = 3.75 m - 1.6 m,
Caution: the question asks for change in speed.

d) Think about the time it takes to make the 2 trips. The change in speed depends on the length of time that the acceleration acts. The kinematic formula
Vf = Vi + a*t
describes the changing velocity over time.
If the time between top and bottom is the same whether going up or down, then the change in speed should be the same.