v = 5 - 9.8 t
0 = 5 - 9.8 t
t = 5/9.8 = .51 seconds to top where v = 0
average speed going up = 5/2 = 2.5 m/s
h = 1.5 + 2.5 * .51 = 2.78 meters
(a) no way
(b)The same speed it will have if dropped from the top of the wall and landing on a platform 1.5 m high
3.95 - 1.5 = 2.45 meters down
2.45 = (1/2)g t^2
2.45 = 4.9 t^2
t = .707 seconds
v = g t = 9.8*.707 = 6.93 meters/second initial speed up
c and d
change in speed /change in time = g
An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.50 m above the ground.
(a) Will the rock reach the top of the wall?
(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?
(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5.00 m/s and moving between the same two points.
(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? Explain physically why it does or does not agree.
1 answer
2 answers