Problem Solving tips for free fall1. Acceleration is always -9.81 m/s/s2. Even at the top3. v at the top is zero

Iff start and end at same elevation4. 1/2 time up, 1/2 time down5. v = -uMake a v vs. time graph for an object thrown straight up from leaving my hand, to landing back in my hand(draw axes, endpoints)

Time

+ V

0 m/s

- V

Bottom (start)

Top

Bottom (end)

Velocity vs. time

Example 1 - An object is launched straight up with a velocity of +33 m/s, and strikes the ground at the same elevation from which it is launched. Use the acceleration of gravity to be -9.81 m/s/s, and neglect air friction1.When is it at the top? 2.How high does it go at the highest? 3.What total time is it in the air?4.What is its velocity at an elevation of 45 m?

3.4 s, 56 m, 6.7 s, +14 m/s OR -14 m/s

Example 2 - A piece falls off a rocket that is 27.0 m above the ground when the rocket is moving upwards at 21.0 m/s. Neglect air friction.

1. Describe to your neighbor what the piece will do2. To what height does the piece rise above its release point

before coming down?3. What is the velocity of impact of the piece with the ground?4. What time does the piece take to reach the ground?

22.5 m, -31.2 m/s, 5.32 s

Whiteboards:Free Fall

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A steel marble falls from the roof of a building that is 43 m tall. In what time does it reach the ground, and what is its velocity of impact? ignore air friction. Use 9.81 as g, and make down negative.

s = -43 m, u = 0 m/s, v = ??, a = -9.81 m/s/s, t = ??s = ut + 1/2at2

.:. t = 2.9608 s 3.0 sv = u + at.:. v = -29.0458 m/s -29 m/s

3.0 s -29 m/s

A baseball is popped up and lands on top of a light tower that is 23.8 m above the elevation of the bat after having been in the air for 4.75 seconds.

a) What is the initial upward velocity the ball has?b) With what velocity does it strike the tower?c) What is the greatest height the ball reaches?Use 9.81 as g, and neglect air friction, down is -

s = +23.8 m, u = ??, v = ??, a = -9.81 m/s/s, t = 4.75 ss = ut + 1/2at2

.:. u = 28.309 m/s 28.3 m/sv = u + at.:. v = -18.2882 m/s -18.3 m/ss = ??, u = 28.309 m/s, v = 0, a = -9.81 m/s/s, t = ??v2 = u2 + 2as.:. s = 40.8468 m 40.8 m

28.3 m/s, -18.3 m/s, 40.8 m

A pop fly remains airborne for 8.42 seconds and is caught at bat level. Ignore air friction.

1. What time did it spend going up?2. What was its initial velocity? Its final?3. What maximum height does the ball reach?4. What was the velocity at an elevation of 60. m?5. Why are there two answers to the previous?6. What is the ball’s velocity and displacement at 5.05 s?

+41.3 m/s, final is -41.3 m/s, 86.9 m, ±23.0 m/s, going up/down, 83.5 m