Question

An attacker at the base of a castle wall 3.55 m high throws a rock straight up with speed 4.50 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 4.50 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?

Yes or No    

Answer 1

(a)
initial speed of the rock (u) = 4.5 m/s
acceleration (a) =-g = -9.81 m/s²
using the kinematic equation
v² = u² + 2aS
at the maximum position , velocity becomes zero therefore
0 = 4.5² +2(-9.81)S
S = 1.03 m
Since the rock cover only 1.03 m while the remaining height of the wall is R = 1.95 m
Therefore rock will not reach to the top of the wall.
(b)
Let us consider that rock will cover the rest of the height R = 1.95 m
let us say its initial velocity = U
Now using kinematic equation
v² = U² +2aR
0 = U² +2(-9.81)*1.95
U = 5.425 m/s
hence rock should have initial velocity 5.425 m/s to reach up to top of the wall .
(c)
If it is throw downward then
acceleration (a) = g = 9.81 m/s²
Now the initial velocity of rock (u) = 4.5 m/s
Using the kinematic equation
v² = u² + 2aR
v² = 4.5² +2(9.81)*1.95
v = 7.65 m/s
Hence the change in the speed = v - u = 7.65 -4.5 = 3.15 m/s
Hence the change in the speed would be 3.15 m/s
(d)
No there magnitude will be different.

R-H-h-3.55-1.6 R-1.95 m H-3.55 mWall u-4.5 m/s h 1.6 m