An attacker at the base of a castle wall 3.60 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 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 8.00 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? Explain physically why it does or does not agree.

Explanation / Answer

Height covered by stone =U^2 /2g =8^2/2*9.8 =3.27 m

As it was thrown from 1.7 m

Total height =3.27+1.7 =4.97 m

a) So stone will reach top of wall

b)Speed at top of wall

Top of wall is at =3.6-1.7 =1.9 m high from the point of throw

So speed at 1.9 m by energy conservation

V^2 = U^2 -2gh =8*8-2*9.8*1.9 =26.76

V at top of wall =5.17 m/sec (ans)

c) When stone is thrown down

V^2 =U^2 +2gh =8*8+2*9.8*1.9 =101.24

V= 10.06 m/sec

So

Change in speed =10.06 - 8 =2.06 m/sec (ans)

d) Change in speed when thrown upwards =8-5.17 =2.83 m/sec

It is not equal as dependance of speed & height depends on V^2 & h and direction of gravity .

And gravity direction is changed in both case so change in speed is not equal in both cases

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