1- A rock is thrown directly upward from the edge of the roof of a building that
ID: 1532468 • Letter: 1
Question
1- A rock is thrown directly upward from the edge of the roof of a building that is 33 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what velocity was the rock thrown? What is the maximum height of the rock above theground? How long does the rock take to get to its maximum height? What is the rock’s velocity just before hitting the ground? Take your y-axis origin to be at the ground with positive upward.
SORT – from roof upward to the ground
Initial position, yo = m
Initial velocity, Vyo = ?
Acceleration, ay = m/s2
Final Position, yf = 0.0 (at the ground)
Final Velocity, vfy = ?
Elapsed time, t = sec
2- SORT – from roof to MAX HEIGHT above ground
Initial position, yo = 33 m (origin at ground)
Initial velocity, Vyo = ?
Acceleration, ay = m/s2
Final Position, yfmaxheight = ? (at max height)
Final Velocity, vfymaxheight = m/s (at max height)
Elapsed time, tmaxheight = ? sec (at max height)
3- The initial velocity of the rock will be ______ and the final velocity of the rock just before hitting the ground will be _____
The initial velocity of the rock
Final velocity of the rock just before hitting the ground
4-
GUESTIMATE:
If another object were dropped (Voy=0) from another building and took 4 seconds to fall from that drop point, we could find the height at which the object was dropped and the velocity of the object just before hitting the ground. These values would bound the maximum height the rock thrown upward could have and also the velocity at which the rock hits the ground. Round the acceleration to the nearest integer. Find the height an object is dropped from if it takes 4 seconds to hit the ground and find the velocity just before it hits the ground. A calculator is not necessary for these calculations.
Height of object if it is dropped and takes 4 seconds to hit the ground and the acceleration is rounded to nearest integer, Yflimit = m
Velocity of object just before hitting the ground if it is dropped and takes 4 seconds to hit the ground and the acceleration is rounded to nearest integer, VfLimit = m/s
5-
SOLVE – for Initial Velocity
The initial velocity of the rock, Voy = ____ m/s DO NOT PUT UNITS IN THE ANSWER BOX
6-
SOLVE – for Final Velocity
The velocity of the rock just before hitting the ground, Vfy = ____ m/s DO NOT PUT UNITS IN THE ANSWER BOX
7-
SOLVE – for MAX height
The max height of the rock above the ground, yfmaxheight = ____ m DO NOT PUT UNITS IN THE ANSWER BOX
8-
SOLVE – for Time to MAX height
The time to get to the max height, tfmaxheight = ____ sec DO NOT PUT UNITS IN THE ANSWER BOX
negative
zero
positive
- A. B. C.The initial velocity of the rock
- A. B. C.Final velocity of the rock just before hitting the ground
4-
GUESTIMATE:
If another object were dropped (Voy=0) from another building and took 4 seconds to fall from that drop point, we could find the height at which the object was dropped and the velocity of the object just before hitting the ground. These values would bound the maximum height the rock thrown upward could have and also the velocity at which the rock hits the ground. Round the acceleration to the nearest integer. Find the height an object is dropped from if it takes 4 seconds to hit the ground and find the velocity just before it hits the ground. A calculator is not necessary for these calculations.
Height of object if it is dropped and takes 4 seconds to hit the ground and the acceleration is rounded to nearest integer, Yflimit = m
Velocity of object just before hitting the ground if it is dropped and takes 4 seconds to hit the ground and the acceleration is rounded to nearest integer, VfLimit = m/s
5-
SOLVE – for Initial Velocity
The initial velocity of the rock, Voy = ____ m/s DO NOT PUT UNITS IN THE ANSWER BOX
6-
SOLVE – for Final Velocity
The velocity of the rock just before hitting the ground, Vfy = ____ m/s DO NOT PUT UNITS IN THE ANSWER BOX
7-
SOLVE – for MAX height
The max height of the rock above the ground, yfmaxheight = ____ m DO NOT PUT UNITS IN THE ANSWER BOX
8-
SOLVE – for Time to MAX height
The time to get to the max height, tfmaxheight = ____ sec DO NOT PUT UNITS IN THE ANSWER BOX
A.negative
B.zero
C.positive
Explanation / Answer
(1)
The vertical position of the rock is given by the equation for free fall
y(t) = y0 + v0t-1/ 2 gt2 . we can solve this for v0:
v0= ((1/2)gt2-y0)/t Putting in y0 = 33, t = 4.00 s, g = 9.81 m/s, we get v0 = 11.35m/s
time = u/g
= 11.35/9.8
= 1.158s
maximum height = 33+11.352/2*9.8
= 39.57m
velocity before hittin ground
v2 - u2 = 2x9.8x33
v2 = 11.352 +2x9.8x33
v = 27.85 m/s
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