3.5 Free fall (Page 5/7)

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Problems

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, 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 $y_{0} = 0$ .

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Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, 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. $\begin{matrix} y = −8.23 m \\ v_{1} = − 18.9 m/s \end{matrix}$ ;
b. $\begin{matrix} y = −18.9 m \\ v_{2} = 23.8 m/s \end{matrix}$ ;
c. $\begin{matrix} y = −32.0 m \\ v_{3} = − 28.7 m/s \end{matrix}$ ;
d. $\begin{matrix} y = −47.6 m \\ v_{4} = − 33.6 m/s \end{matrix}$ ;
e. $\begin{matrix} y = −65.6 m \\ v_{5} = − 38.5 m/s \end{matrix}$

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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?

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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. Knowns: $a = − 9.8 {m/s}^{2} v_{0} = −1.4 m/s t = 1.8 s y_{0} = 0 m$ ;
b. $y = y_{0} + v_{0} t - \frac{1}{2} g t^{2} y = v_{0} t - \frac{1}{2} g t = −1.4 m / s (1.8 sec) - \frac{1}{2} (9.8) {(1.8 s)}^{2} = −18.4 m$ and the origin is at the rescuers, who are 18.4 m above the water.

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Unreasonable results A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 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 a time is the dolphin in the air? Neglect any effects resulting from his size or orientation.

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A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

a. $v^{2} = v_{0}^{2} - 2 g (y - y_{0}) y_{0} = 0 v = 0 y = \frac{v_{0}^{2}}{2 g} = \frac{{(4.0 m / s)}^{2}}{2 (9.80)} = 0.82 m$ ; b. to the apex $v = 0.41 s$ times 2 to the board = 0.82 s from the board to the water $y = y_{0} + v_{0} t - \frac{1}{2} g t^{2} y = −1.80 m y_{0} = 0 v_{0} = 4.0 m / s$ $−1.8 = 4.0 t - 4.9 t^{2} 4.9 t^{2} - 4.0 t - 1.80 = 0$ , solution to quadratic equation gives 1.13 s; c. $\begin{matrix} v^{2} = v_{0}^{2} - 2 g (y - y_{0}) y_{0} = 0 v_{0} = 4.0 m / s y = −1.80 m \\ v = 7.16 m / s \end{matrix}$

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(a) Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of 8.00 m/s. (b) How long a time would it take to reach the ground if it is thrown straight down with the same speed?

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A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. How long a time does he have to get out of the way if the shot was released at a height of 2.20 m and he is 1.80 m tall?

Time to the apex: $t = 1.12 s$ times 2 equals 2.24 s to a height of 2.20 m. To 1.80 m in height is an additional 0.40 m. $\begin{matrix} y = y_{0} + v_{0} t - \frac{1}{2} g t^{2} y = −0.40 m y_{0} = 0 v_{0} = −11.0 m / s \\ y = y_{0} + v_{0} t - \frac{1}{2} g t^{2} y = −0.40 m y_{0} = 0 v_{0} = −11.0 m / s \\ - 0.40 = −11.0 t - 4.9 t^{2} or 4.9 t^{2} + 11.0 t - 0.40 = 0 \end{matrix}$ .
Take the positive root, so the time to go the additional 0.4 m is 0.04 s. Total time is $2.24 s + 0.04 s = 2.28 s$ .

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You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.0 m. How much additional time elapses before the ball passes the tree branch on the way back down?

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A kangaroo can jump over an object 2.50 m high. (a) Considering just its vertical motion, calculate its vertical speed when it leaves the ground. (b) How long a time is it in the air?

a. $\begin{matrix} v^{2} = v_{0}^{2} - 2 g (y - y_{0}) y_{0} = 0 v = 0 y = 2.50 m \\ v_{0}^{2} = 2 g y \Rightarrow v_{0} = \sqrt{2 (9.80) (2.50)} = 7.0 m / s \end{matrix}$ ; b. $t = 0.72 s$ times 2 gives 1.44 s in the air

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Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105.0 m. He can’t see the rock right away, but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

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There is a 250-m-high cliff at Half Dome in Yosemite National Park in California. Suppose a boulder breaks loose from the top of this cliff. (a) How fast will it be going when it strikes the ground? (b) Assuming a reaction time of 0.300 s, how long a time will a tourist at the bottom have to get out of the way after hearing the sound of the rock breaking loose (neglecting the height of the tourist, which would become negligible anyway if hit)? The speed of sound is 335.0 m/s on this day.

a. $v = 70.0 m / s$ ; b. time heard after rock begins to fall: 0.75 s, time to reach the ground: 6.09 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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