An attacker at the base of a castle wall 3.65m high throws a rockstraight up wit

Question

An attacker at the base of a castle wall 3.65m high throws a rockstraight up with speed 7.40 m/s at a height of 1.55 m aboveground.
A) will the rock reach the top of the wall
B) If so, what is its speed at the top? If not, what initialspeed must it have to reach the top?
C) Find the change in speed of the downward-moving rock agreewith the magnitude of the speed change of the rock moving upwardbetween the same elevations? Explain physically why it does or doesnot agree.
I GET STUCK AT EVERYWHERE. If someone knows how to do thisproblem, please help!
A) will the rock reach the top of the wall
B) If so, what is its speed at the top? If not, what initialspeed must it have to reach the top?
C) Find the change in speed of the downward-moving rock agreewith the magnitude of the speed change of the rock moving upwardbetween the same elevations? Explain physically why it does or doesnot agree.
I GET STUCK AT EVERYWHERE. If someone knows how to do thisproblem, please help!

Explanation / Answer

A) will the rock reach the top of the wall
maximum height=u*u/2g=7.4*7.4/2*9.8=2.793m from the position ofthrow
hence maximum height=4.34 m
hence the rock reaches the maximumheight.


B) If so, what is its speed at the top? If not, what initialspeed must it have to reach the top?
u=7.4m/sec
a=g=-9.8m/sec2
v=?
s=2.1m
using v*v=u*u-2gh
hence v= 3.68m/sec

C) Find the change in speed of the downward-moving rock agreewith the magnitude of the speed change of the rock moving upwardbetween the same elevations? Explain physically why it does or doesnot agree.
speed change is the same .
it does agree . if there is no airresistance .
it does not agree when there is air no airresistance

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