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an attacker at the base of a castle wall 3.80 m high throws a rock straight up w

ID: 1405176 • Letter: A

Question

an attacker at the base of a castle wall 3.80 m high throws a rock straight up with speed 5.00 m/s from height of 1.4 m above the ground. what is the speed of the rock at the top of the wall? find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5 m/s and moving between the same two points. an attacker at the base of a castle wall 3.80 m high throws a rock straight up with speed 5.00 m/s from height of 1.4 m above the ground. what is the speed of the rock at the top of the wall? find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5 m/s and moving between the same two points. an attacker at the base of a castle wall 3.80 m high throws a rock straight up with speed 5.00 m/s from height of 1.4 m above the ground. what is the speed of the rock at the top of the wall? find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 5 m/s and moving between the same two points.

Explanation / Answer

lets assume that upwards is positve and downwards is negative .

then initial speed=5 m/s, upwards

acceleration=-9.8 m/s^2, downwards (hence the negative sign)

the highest point that the rock rises upto is where its velocity drops to zero.

so using final speed^2-initial speed^2=2*acceleration*distance

where final speed=0 m/s

initial speed=5 m/s

acceleration=-9.8 m/s^2

we get distance=1.275 m

so maximum height achieved by the rock=1.4+1.275=2.675 m

that means the rock wont rise upto the top of the wall.

part 2:

if a rock is trwon at an initial speed of 5 m/s, let the final speed when the rock reaches ground is v.

then v^2-5^2=2*9.8*3.8

==> v=8.63 m/s

hence the speed at which the rock will hit the ground is 8.63 m/s