Question

An attacker at the base of a castle wall 3.75 m high throws a rock straight up with speed 8.50 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 8.50 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?

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

Answer 1

apply the Kinematic equation of spped as V = u + at

here a = accleration due to gravity = -9.81 m/s^2

so at the top position of height , V = 0

so

0 = 8.5 - 9.81 t

t = 8.5/9.81

t = 0.866 secs

now using another Kinamatic equation

y(t) = Y + ut + 0.5 gt^2

Y(t) = { (1.5) + (8.5 * 0.866)- (0.5* 9.81 * 0.866* 0.866)}

y(t) = 5.18 m

so total height = (3.75 + 1.5) = 5.25 m

so naswer is no. it wont reach top

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b. use g = u/t

t = Vo/9.81

y(t) = 3.75 = -4.9 (v0/9.8)^2 + v0 (v0/9.8) + 1.5

= -1/2*9.8 v^2 + 1/9.8 v^2 + 1.5

= 1/(2*9.8) v0^2 + 1.5

v0 = sqrt(2*9.8 (3.75 - 1.5)) = 6.64 m/s

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c) use S = H + ut - 0.5gt^2

y(t) = 1.5 = -4.9 t^2 - 8.5 t + 3.75

4.9 t^2 + 8.5 t - 2.28 = 0

solivng using quadratic expression we get

t = 0.252

delta v = a t = (8.5- (0.252* 8.85)) = 5.18 m/s

d) It's different because it takes different amount of time to move between the points.

e. the effect of grativation force is not coincinding with the accleration prodcued in the body

and hence the results