An attacker at the base of a castle wall 3.55 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.55 m high throws a rock straight up with speed 8.00 m/s from a height of 1.60 m above the ground.

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

Yes or No    

(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?
___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 8.00 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? Explain physically why it does or does not agree.

Explanation / Answer

height of castle wall, H = 3.55 m
height of attacker from the ground, h = 1.6 m
initial speed of rock, thrown vertically up u = 8 m/s

a. max height of the rock = h'
   using
   2*g*h' = u^2
   h' = 64/2*(9.81) = 3.261 m

   so maximum height = h' + h = 4.861 m

   so the rock doies reach top of the wall

b. speed at top = v
   s = 3.55 - 1.6 = 1.95
   using 2*g*s = u^2 - v^2
   2*9.81*1.95 = 64 - v^2
   v = 5.07 m/s
   speed change, u - v = 2.93 m/s

c. speed at top , u = 8 m/s
   distance s = 1.95 m
   final speed = v
   using
   2gs = v^2 - u^2
   2*9.81*1.95 = v^2 - 64
   v = 10.112 m/s

   sop change in speed = v - u = 2.112 m/s

d. the magnitude of speed change is not the same for both the cases, becuase of the fact that square of the velocity is used in calculation and not the magnitude
   from the equaiton
   2gh = v^2 - u^2

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