Physics, 18.02.2020 17:20 hjohnsonn102010

You toss a racquetball directly upward and then catch it at the same height you released it 2.16 s later. Assume air resistance is negligible.

(a) What is the acceleration of the ball while it is moving upward?

magnitude
direction
(b) What is the acceleration of the ball while it is moving downward?

magnitude
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(c) What is the acceleration of the ball while it is at its maximum height?

magnitude
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(d) What is the velocity of the ball when it reaches its maximum height?

magnitude
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(e) What is the initial velocity of the ball?

magnitude
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(f) What is the maximum height that the ball reaches?

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Answer from: durham953

This question involves the concepts of acceleration due to gravity and the equations of motion.

a) The acceleration of the ball while moving upward is "9.81 m/s²" in "downward direction".

b) The acceleration of the ball while moving downward is "9.81 m/s²" in "downward direction".

c) The acceleration of the ball while it is at the highest point is "9.81 m/s²" in "downward direction".

d) The velocity of the ball at maximum height is "0 m/s".

e) The initial velocity of the ball is "10.6 m/s".

f) The maximum height reached by the ball is "17.16 m".

The acceleration of an object during the vertical motion is always constant and it is equal to the acceleration due to gravity. Therefore,

Magnitude: 9.81 m/s²

Direction: Downward

Magnitude: 9.81 m/s²

Direction: Downward

Magnitude: 9.81 m/s²

Direction: Downward

When the ball reaches the maximum height it stops for a moment to change its direction from upward to downward. Hence at the maximum height, the speed of the ball is 0 m/s.

Using the first equation of motion for the upward part of the motion to find out the initial velocity of the ball:

v_f=v_i+gt

where,

vf = final speed at highest point = 0 m/s

vi = initial speed = ?

g = acceleration due to gravity = 9.81 m/s²

t = time taken for upward motion = 2.16 s/2 = 1.08 s

Therefore,

$0\ m/s=v_i+(9.81\ m/s^2)(1.08\ s)\\\\v_i=10.6\ m/s$

Now, we will use the second equation of motion to find out the maximum height:

$h=v_it+\frac{1}{2}gt^2\\\\h=(10.6\ m/s)(1.08\ s)+\frac{1}{2}(9.81\ m/s^2)(1.08\ s)^2\\\\h=11.44\ m+5.72\ m$

h = 17.16 m

Learn more about equations of motion here:

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The attached picture shows the equations of motion.

Answer from: derioncapers2346

All results below

Explanation:

Vertical Throw

An object being thrown upwards in free air (neglibible friction) has a motion with changing velocity ruled only by the acceleration of gravity that attracts all the objects near the planet to its center or in our case, to the ground level.

The racquet ball is tossed upward and caught 2.16 seconds later at the same height. Let's answer each part:

(a) What is the acceleration of the ball while it is moving upward?

The only acceleration acting on the ball is the acceleration of gravity, regardless of the ball's position or time in the air. Thus:

Magnitude: $9.8\ m/s^2$

Direction: -90° (negative y-axis)

(b) What is the acceleration of the ball while it is moving downward?

Same answer as part (a), the acceleration is constant.

Magnitude: $9.8\ m/s^2$

Direction: -90° (negative y-axis)

Same answer as parts (a) and (b), the acceleration is constant.

Magnitude: $9.8\ m/s^2$

Direction: -90° (negative y-axis)

(d) What is the velocity of the ball when it reaches its maximum height?

The initial speed at which the ball was launched gradually goes to zero as the object goes higher. When the speed finally is 0, the object reaches its maximum height, thus:

V=0 m/s

Direction: Does not apply

(e) What is the initial velocity of the ball?

The flight time is twice the time it takes to reach the maximum height, thus the ball goes up for 2.16/2=1.08 seconds and goes down for 1.08 seconds. The speed it's launched is the same speed is returns back to the launch point and is computed by

$v_f=g.t=9.8\cdot 1.08=10.584\ m/s$

Thus the initial speed is

$v_o=10.584\ m/s$

Direction: +90° (positive y-axis)

(f) What is the maximum height that the ball reaches?

$\displaystyle y=\frac{gt^2}{2}=\frac{9.8\times 1.08^2}{2}=5.72\ m$

The maximum height is 5.72 m

You toss a racquetball directly upward and then catch it at the same height you released it 2.16 s l

Answer from: Quest

hey , the answer is c

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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