Question
Answer the following questions
Show your work. Use g = 9.80 m/s2. Assume no air friction. A golf ball released (from rest) from a height of 1.50 m above a concrete floor bounces back to a height of 1.10 m. If the ball is in contact with the floor for 0.620 ms, what is the average acceleration of the ball while in contact with the floor? A rock is dropped (from rest) from the top of a vertical cliff next to the ocean and the sound of it striking the ocean is heard 3.40 seconds later. The speed of sound is 340 m/s. How high is the cliff? A falling stone (that fell with an initial velocity of zero) takes 0.300 seconds to travel past a window 2.20 m tall. From what height above the top of the window did the stone fall? Solve this problem in two different ways. First, solve the problem by starting with the average velocity of the stone as it passes across the window but note that the average velocity is NOT its velocity at the midpoint of the window but its velocity at the mid time point while crossing the window. Second, solve the problem by constructing two equations, one giving the displacement from the start to the top of the window and the other the displacement from the start to the bottom of the window and then solving simultaneously. Suppose you throw a stone straight upwards with an initial speed of 15.0 m/s. If you throw a second stone straight up 1.00 seconds after the first, with what speed must you throw this stone if it is to hit the first stone at a height of 11.0 m? There are two answers, one realistic and one not. A tardy commuter sprints at a constant speed of 4.60 m/s in an attempt to catch a train that is about to pull away from the station. The commuter runs parallel to the tracks and is still a distance X from the door at the end of the train when the train starts to accelerate from rest at 0.380 m/s2. Find the largest value of X for which the commuter catches the train.
Explanation / Answer
1)
so when it hits the floor it will have a speed = sqrt(2*g*y)
so dv/dt = (sqrt(2*9.81*1.5)-sqrt(2*9.81*1.1))/0.62E-2=125.7 N
2)
time for it hit
h = 1/2 a t^2
t = sqrt(2 h/g)
so time total = tfall + tsound=sqrt(2h/g)+h/340 = 3.4
sqrt(2h/9.81)+h/340 = 3.4
h= 51.74 m
3)
so 0=h + v0 t - 1/2 a t^2
0 = 2.2 + v*0.3 - 0.5*9.81*0.3^2
v=-5.862 m/s
to get that speed v^2 = v0^2 + 2 a y
y = 5.862/(2*9.81)=0.3 m
4) first stone
y = 15 t - 0.5*9.81*t^2
11 = 15 t - 0.5*9.81*t^2
t=1.22 and 1.84
second stone
y = v*(t-1) - 0.5*9.81*(t-1)^2
first answr
11 = v*.22 - 0.5*9.81*.22^2
v= 51 m/s
second answer
11 = v*.84 - 0.5*9.81*.84^2
v=17.2 m/s
5)
x commuter = 4.6 t
x train = 1/2*0.38*t^2 + x
4.6 t = 0.5*0.38*t^2 + x
x = 4.6 t - 0.5*0.38*t^2
has max when dx/dt = 0
dx/dt = 4.6 - 2*0.5*0.38*t =0
t=12.105 s
so X = 27.84 m