An attacker at the base of a castle wall 3.80 m high throws a rock straight up w

Question

An attacker at the base of a castle wall 3.80 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 Yes No is its speed at the top? If not, what initial speed must it have to reach the top? 779 m/'s c) Find the change in speed of a rock thrown n straight down from the top of the wall at an initial speed of 8.00 m/s and moving between the same two points te date results to a Your response is within 10% of the correct value. This may be de to roundoff error or you could have a mistake in your calculator. Carry out all least four digit accuracy to roinimize roundoff error. 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? O Yes

Explanation / Answer

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

And it was thrown from 1.7 m

Total height =3.27+1.7 =4.97 m

a)Therefore stone will reach top of wall

b)Speed at top of wall

Top of wall is at =3.8-1.7 =2.1 m high from the point of throw

So speed at 2.1 m by energy conservation

V^2 = U^2 -2gh =8*8-2*9.8*2.1 =22.84

So V at top of wall =4.78 m/sec (ans)

c) When stone is thrown down

V^2 =U^2 +2gh =8*8+2*9.8*2.1 =105.16

V= 10.25 m/sec

So

Change in speed =10.25 - 8 =2.25 m/sec (ans)

d) Change in speed when thrown upwards =8-4.78 =3.22 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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