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A hot-air balloon rises from ground level at a constant velocity of 3.0 m/s. One minute after liftoff, a sandbag is dropped accidentally from the balloon. Calculate (a) the time it takes for the sandbag to reach the ground and (b) the velocity of the sandbag when it hits the ground.

a. t = 6.37 s taking the positive root;
b. v = 59.5 m/s

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(a) A world record was set for the men’s 100-m dash in the 2008 Olympic Games in Beijing by Usain Bolt of Jamaica. Bolt “coasted” across the finish line with a time of 9.69 s. If we assume that Bolt accelerated for 3.00 s to reach his maximum speed, and maintained that speed for the rest of the race, calculate his maximum speed and his acceleration. (b) During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race?

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An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b) Determine the final velocity at which the object hits the ground. (c) Determine the distance traveled during the last second of motion before hitting the ground.

a. y = 4.9 m ;
b. v = 38.3 m/s ;
c. −33.3 m

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A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms ( 8.00 × 10 5 s ) (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

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An object is dropped from a roof of a building of height h . During the last second of its descent, it drops a distance h /3. Calculate the height of the building.

h = 1 2 g t 2 , h = total height and time to drop to ground
2 3 h = 1 2 g ( t 1 ) 2 in t – 1 seconds it drops 2/3 h
2 3 ( 1 2 g t 2 ) = 1 2 g ( t 1 ) 2 or t 2 3 = 1 2 ( t 1 ) 2
0 = t 2 6 t + 3 t = 6 ± 6 2 4 · 3 2 = 3 ± 24 2
t = 5.45 s and h = 145.5 m. Other root is less than 1 s. Check for t = 4.45 s h = 1 2 g t 2 = 97.0 m = 2 3 ( 145.5 )

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Challenge problems

In a 100-m race, the winner is timed at 11.2 s. The second-place finisher’s time is 11.6 s. How far is the second-place finisher behind the winner when she crosses the finish line? Assume the velocity of each runner is constant throughout the race.

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The position of a particle moving along the x -axis varies with time according to x ( t ) = 5.0 t 2 4.0 t 3 m. Find (a) the velocity and acceleration of the particle as functions of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position.

a. v ( t ) = 10 t 12 t 2 m/s, a ( t ) = 10 24 t m/s 2 ;
b. v ( 2 s ) = −28 m/s, a ( 2 s ) = −38 m/s 2 ; c. The slope of the position function is zero or the velocity is zero. There are two possible solutions: t = 0, which gives x = 0, or t = 10.0/12.0 = 0.83 s, which gives x = 1.16 m. The second answer is the correct choice; d. 0.83 s (e) 1.16 m

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A cyclist sprints at the end of a race to clinch a victory. She has an initial velocity of 11.5 m/s and accelerates at a rate of 0.500 m/s 2 for 7.00 s. (a) What is her final velocity? (b) The cyclist continues at this velocity to the finish line. If she is 300 m from the finish line when she starts to accelerate, how much time did she save? (c) The second-place winner was 5.00 m ahead when the winner started to accelerate, but he was unable to accelerate, and traveled at 11.8 m/s until the finish line. What was the difference in finish time in seconds between the winner and runner-up? How far back was the runner-up when the winner crossed the finish line?

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In 1967, New Zealander Burt Munro set the world record for an Indian motorcycle, on the Bonneville Salt Flats in Utah, of 295.38 km/h. The one-way course was 8.00 km long. Acceleration rates are often described by the time it takes to reach 96.0 km/h from rest. If this time was 4.00 s and Burt accelerated at this rate until he reached his maximum speed, how long did it take Burt to complete the course?

96 km/h = 26.67 m/s, a = 26.67 m/s 4.0 s = 6.67 m/s 2 , 295.38 km/h = 82.05 m/s, t = 12.3 s time to accelerate to maximum speed
x = 504.55 m distance covered during acceleration
7495.44 m at a constant speed
7495.44 m 82.05 m/s = 91.35 s so total time is 91.35 s + 12.3 s = 103.65 s .

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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