An attacker at the base of a castle wall 3.90 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.90 m high throws a rock straight up with speed 5.50 m/s from a height of 1.60 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?
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 5.50 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?

YesNo    

(e) Explain physically why it does or does not agree.

Explanation / Answer

Gvien that

An attacker at the base of a castle wall (h) = 3.90 m high

Throws a rock straight up with speed (v) = 5.50 m/s from a height of 1.60 m above the ground.

Here we consider a =-g =9.81m/s2

After reaching to a height final velocity v(t) =0

v(t) =u+at ===> 0=5.50-9.81(t) ===> t =5.50/9.81 =0.560s

Therefore y(t) =y(o) +ut+(1/2)gt2 ===> y(t) =1.60+5.50(0.560)-0.5*9.81(0.560)2===>+1.60+3.08-1.538=3.142m

So, the wall reaches the top of the wall

b)

The speed at the top of wall is given by

t = vi/9.8
y(t) = 3.90 = -4.9 (vi/9.81)2 + v0i(vi/9.81) + 1.60 = -1/2*9.81 v^2 + 1/9.81 v^2 + 1.60 = 1/(2*9.81) vi^2 + 1.60
vi = sqrt(2*9.81 (3.90 - 1.60)) = 6.717m/s

c)

The inital speed of the projection is vo
vf be the speed at distance h below.
vf2 = vo2 + 2gh ...(2)
v = sqrt[ 5.502 + 2 * 9.81(3.90 - 1.60) ]
= 8.68 m/s.

d)

The change in speed is given by

vf -vi =5.50-6.717 =-1.217m/s for answer (b)

Now the change apeed for answer (c) is given by =8.68-6.717=1.963m/s

Therefore the change in speed is different for different elevations.

Therefore it does not agree.

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