Ask question

Ask question

All categories

SVETLANKA909090 [29]

3 years ago

14

A baseball is thrown by the center fielder (from shoulder level) to home plate where it is caught (on the fly at eye level) by t

he catcher. At what point does the magnitude of the vertical component of velocity have its maximum value? (air resistance is negligible)

1 answer:

marishachu [46]3 years ago

8 0

At the highest point of the trajectory the vertical component will have its zero velocity, and the descent caused by the force of gravity will begin.

Since the ball is thrown with a certain speed, the vertical component reaches its highest point (upwards), until returning to the receiver who will receive the ball with the same vertical component but in the opposite direction (downwards).

Therefore the vertical component will have its highest value at launch.

You might be interested in

A physics student with too much free time drops a watermelon from a roof of a building, hears the sound of the watermelon going

tatiyna

Answer:

28.6260196842 m

Explanation:

Let h be the height of the building

t = Time taken by the watermelon to fall to the ground

Time taken to hear the sound is 2.5 seconds

Time taken by the sound to travel the height of the cliff = 2.5-t

Speed of sound in air = 340 m/s

For the watermelon falling

$s=ut+\frac{1}{2}at^2\\\Rightarrow h=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow h=\frac{1}{2}\times 9.81\times t^2$

For the sound

Distance = Speed × Time

$\text{Distance}=340\times (2.5-t)$

Here, distance traveled by the stone and sound is equal

$\frac{1}{2}\times 9.81\times t^2=340\times (2.5-t)\\\Rightarrow 4.905t^2=340\times (2.5-t)\\\Rightarrow t^2=\frac{340}{4.905}(2.5-t)\\\Rightarrow t^2+69.3170234455t-173.292558614=0$

$t=\frac{-69.31702\dots +\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1},\:t=\frac{-69.31702\dots -\sqrt{69.31702\dots ^2-4\cdot \:1\cdot \left(-173.29255\dots \right)}}{2\cdot \:1}\\\Rightarrow t=2.4158\ s\ or\ -71\ seconds$

The time taken to fall down is 2.4158 seconds

$h=\frac{1}{2}\times 9.81\times 2.4158^2=28.6260196842\ m$

Height of the buidling is 28.6260196842 m

7 0

4 years ago

Please help, I really don't know

zubka84 [21]

circular motion.

cent acc = r omega^2 ... omega is ang vel ... omega=2pi/T ,,,

9.8=rx(2pi/T)^2

if r is known, solve for T

7 0

4 years ago

After giving an intense performance, a confused and disoriented ﬂautist has wandered onto the motorway! They are playing a con

leva [86]

This is a Doppler effect. Generally, if you move to a frequency source, you would detect an increase in frequency and when you move away from a source you would detect a decrease.

For this question, before you pass them, you are actually approaching them, so you would hear a higher frequency than the constant 300 Hz they are playing at.

Using the condensed formula:

f ' = ((v <u>+</u> vd)/(v <u>+</u> vs)) * f

Where: vd = Velocity of the detector.
   vs = Velocity of the frequency source.
   v = Velocity of sound in air.
   f ' = Apparent frequency.
   f = Frequency of source.

v = 343 m/s, vd = detector = 27.8 m/s, vs = velocity of the source =0. (the flautists are not moving).
f = 300 Hz.

There would be an overall increase in frequency, so we maintain a plus at the numerator and a minus at the denominator.

f ' = ((v + vd)/(v - vs)) * f

f ' = ((343+ 27.8)/(343 - 0)) * 300
   = (370.8/343)* 300 = 324.3

Therefore frequency before passing them = 324.3 Hz.

Cheers.

4 0

3 years ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.46 m. The mug

Tresset [83]

Answer:

(a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

Explanation:

Given that,

Height of the counter = 1.46 m

Distance = 0.80 m

We need to calculate the time

Using equation of motion

$s_{y}=ut+\dfrac{1}{2}gt^2$

$s_{y}=\dfrac{1}{2}gt^2$

Put the value into the formula

$1.46=\dfrac{1}{2}\times9.8\times t^2$

$t^2=\dfrac{1.46\times 2}{9.8}$

$t=\sqrt{\dfrac{1.46\times2}{9.8}}$

$t=0.545\ sec$

Here, horizontal velocity is constant

(a). We need to calculate the velocity

Using formula of velocity

$v_{x}=\dfrac{d}{t}$

Put the value into the formula

$v_{x}=\dfrac{0.80}{0.545}$

$v_{x}=1.46\ m/s$

(b). We need to calculate the final velocity

Using equation of motion

$v_{f}^2=u^2+2as$

Put the value into the formula

$v_{f}^2=0+2\times9.8\times1.46$

$v_{f}=\sqrt{28.616}$

$v_{f}=5.34\ m/s$

The velocity is 5.34 m/s downward.

We need to calculate the direction

Using formula of direction

$\tan\theta=\dfrac{v_{y}}{v_{x}}$

$\theta=\tan^{-1}(\dfrac{5.34}{1.46})$

$\theta=74.7^{\circ}$

Hence, (a). The horizontal velocity is 1.46 m/s.

(b). The direction of the mug's velocity just before it hit the floor is 74.7° below the horizontal.

5 0

4 years ago

The earth is rotating on its axis. It will continue to rotate unless acted upon by an outside force. This is an example of Newto

mojhsa [17]

It is an example of Newton’s first law.

7 0

3 years ago

Read 2 more answers