1- A rock is thrown downward from the top of a 41.2-m-tall tower with an initial
ID: 250343 • Letter: 1
Question
1- A rock is thrown downward from the top of a 41.2-m-tall tower with an initial speed of 15 m/s. Assuming negligible air resistance, what is the speed of the rock just before hitting the ground?
2- A gymnast jumps straight up, with her center of mass moving at 3.98 m/s as she leaves the ground. How high above this point is her center of mass at the following times? (Ignore the effects of air resistance, and assume the initial height of her center of mass is at y = 0.)
0.500
3- An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 4.00 m/s from a height of 1.40 m above the ground.
(a) Will the rock reach the top of the wall?
YesNo
(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 4.00 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.
0.500
3- An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 4.00 m/s from a height of 1.40 m above the ground.
(a) Will the rock reach the top of the wall?
YesNo
(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 4.00 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
1)
Using equation, v2 = u2 + 2gh
=> v2= 152 +2x9.8x41.2
=>v = 31.13 m/sec
2)
. For upward motion of ball using ,
s= ut - 1/2gt2 ,
at t= .100 , s= 3.98x .1 - 1/2 x 9.8 x .12= .349m ,
at t=.2 ,s= 3.98x .2 - 1/2 x 9.8 x .22= 0.6 m,
at t=.3, s=3.98x .3 - 1/2 x 9.8 x .32= 0.753,
at t=.4 , s=3.98x .4 - 1/2 x 9.8 x .42= 0.808m ,
at t=.5 s=3.98x .5 - 1/2 x 9.8 x .52= 0.765m
3)
. Let maximum height reached by the rock is h, at max height velocity reaches zero, so
using v2 = u2 - 2gh
0= 42 - 2x 9.8 x h
h= .81 m (maximum height reached)
a. as the wall is 2.20 m up from the point of projection so the rock can not reach the top of wall.
b)
Let the intial velocity be u so that it reaches top of wall,
v2 = u2 - 2gh
u = sqrt( 2gh)
u=2x 9.8x 2.2= 6.57 m/s
c.)
let final velocity when rock is thrown downward is v , then v2 = u2 + 2gh ==> v2 = 42 + 2x9.8x2.2= 7.68 m/sec
change in spped at point of projection= 7.68-4= 3.68 m/sec
d).
Yes change in speed is same for upward and downward motion
e.)
as acceleration is same, it means speed changes constantly. it slows in upward motion by the same amount in same time interval as gain in spped in upward motion in same time interval.
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