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An object that is thrown straight up falls back to Earth. This is one-dimensional motion. (a) When is its velocity zero? (b) Does its velocity change direction? (c) Does the acceleration due to gravity have the same sign on the way up as on the way down?
Suppose you throw a rock nearly straight up at a coconut in a palm tree, and the rock misses on the way up but hits the coconut on the way down. Neglecting air resistance, how does the speed of the rock when it hits the coconut on the way down compare with what it would have been if it had hit the coconut on the way up? Is it more likely to dislodge the coconut on the way up or down? Explain.
If an object is thrown straight up and air resistance is negligible, then its speed when it returns to the starting point is the same as when it was released. If air resistance were not negligible, how would its speed upon return compare with its initial speed? How would the maximum height to which it rises be affected?
The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration due to gravity being the same, how many times higher could a safe fall on the Moon be than on Earth (gravitational acceleration on the Moon is about 1/6 that of the Earth)?
How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (gravitational acceleration on the Moon is about 1/6 of $g$ on Earth)?
Assume air resistance is negligible unless otherwise stated.
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be ${y}_{0}=0$ .
(a) ${y}_{1}=6\text{.}\text{28 m}$ ; ${v}_{1}=\text{10}\text{.}\text{1 m/s}$
(b) ${y}_{2}=\text{10}\text{.}\text{1 m}$ ; ${v}_{2}=5\text{.}\text{20 m/s}$
(c) ${y}_{3}=11\text{.}\mathrm{5\; m}$ ; ${v}_{3}=0\text{.300 m/s}$
(d) ${y}_{4}=10\text{.4 m}$ ; ${v}_{4}=-4\text{.60 m/s}$
Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.
A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?
${v}_{0}=4\text{.}\text{95 m/s}$
A rescue helicopter is hovering over a person whose boat has sunk. One of the rescuers throws a life preserver straight down to the victim with an initial velocity of 1.40 m/s and observes that it takes 1.8 s to reach the water. (a) List the knowns in this problem. (b) How high above the water was the preserver released? Note that the downdraft of the helicopter reduces the effects of air resistance on the falling life preserver, so that an acceleration equal to that of gravity is reasonable.
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s. (a) List the knowns in this problem. (b) How high does his body rise above the water? To solve this part, first note that the final velocity is now a known and identify its value. Then identify the unknown, and discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking units, and discuss whether the answer is reasonable. (c) How long is the dolphin in the air? Neglect any effects due to his size or orientation.
(a) $a=-9\text{.}{\text{80 m/s}}^{2}$ ; ${v}_{0}=\text{13}\text{.}\text{0 m/s}$ ; ${y}_{0}=\text{0 m}$
(b) $v=0\text{m/s}$ . Unknown is distance $y$ to top of trajectory, where velocity is zero. Use equation ${v}^{2}={v}_{0}^{2}+2a\left(y-{y}_{0}\right)$ because it contains all known values except for $y$ , so we can solve for $y$ . Solving for $y$ gives
Dolphins measure about 2 meters long and can jump several times their length out of the water, so this is a reasonable result.
(c) $2\text{.}\text{65 s}$
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