a small stone is dropped from the top of a building and strikes the ground 2 sec
ID: 2171736 • Letter: A
Question
a small stone is dropped from the top of a building and strikes the ground 2 seconds later.a) determine the height of the building and velocity of the when the stone strikes the ground.
b) the same stone is dropped from a different building and takes 4 seconds to land on the ground. Determine the height of the building and the velocity of the stone when the stone strikes the ground.
c)compare the impact speeds for these two stones. how much faster is the second stone at impact? can you explain why it is this much faster?
d)compare the distances traveled by these two stones. how much farther does the second stone travel?
Explanation / Answer
To find the horizontal displacement at .0 s: d = vit + (0.5)at2 where vi = 2.5, t = 2.0 and a = 0 . d = 10 m To find the vertical displacement at 4.0 s: d = vit + (0.5)at2 where vi = 0, t = 4.0 and a = -0.75 m/s2. d = -6 m The position of the fleck at 4.0 s is 10 m to the right and , 6 m down from the point the acceleration started. For solutions to all the problems on this page click here. B28. To find the magnitude of the velocity of the boulder leaving the cliff, we apply vf2 = vi2 + 2ad vf2 = 0 2 + 2(3.06)(35.0) vf = 14.64 m/s horizontal vertical distance ? 45.0 m acceleration 0 9.81 m/s2 initial velocity 14.64cos19.0° 14.64sin19.0° final velocity time t t In the vertical component, applying d = vit + (0.5)at2 45.0 = (14.64sin19.0°) t + (0.5)(9.81)t2 Solving the quadratic for the positive root: t = 2.58 s The boulder is in the air for 2.58 s In the horizontal, applying d = vit + (0.5)at2 d = (14.64cos19.0°)(2.58) + 0 d = 35.7 m The boulder lands 35.7 m from the base of the cliff.
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