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2 years ago

A volleyball serve was in the air for 2.2 seconds before it landed untouched in the far corner of the

Physics

2 answers:

RUDIKE [14]2 years ago

8 0

We are only interested in the vertical motion of the ball.

The ball remains in air for t=2.2 s, so we can say that it reaches its maximum height in t=1.1 s (half the time) before falling down. This is an uniformly accelerated motion with constant acceleration g=9.81 m/s^2, so the maximum height reached by the balls is given by:

$S=\frac{1}{2}gt^2 = \frac{1}{2}(9.81 m/s^2)(1.1s)^2=5.93 m$

lina2011 [118]2 years ago

3 0

Given that air resistance is negative.
Hence the values are: 2.2 s G = 9.8 m/s2
Solution
Velocity = vf-vi = 0 – vi = -vi
V = 9.8m/s2 x 2.2 s = 21.56 m/ s (cancel one s out)
Vi = 21.56 m/s

Then (s) displacement is
S=v x t
Average velocity = (21.56 m/s + 0 / 2)
Average velocity = 10.78 m/s
Time = 2.2 s

S = 10.78 m/s x 2.2 s (cancel s)
S = 23.716 m

Therefore the ball was at 23.716 meters in the air.

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For metalloids on the periodic table, how do the group number and the period number relate?

lapo4ka [179]

Im guessing it's (a) since the numbers go in chronological order and you read the periodic table left to right

3 0

2 years ago

Read 2 more answers

Ariel dropped a golf ball from her second story window. The ball starts from rest and hits the sidewalk 3.5 s later with a veloc

Aleks [24]

Answer:

By using the acceleration formula,

$a = \frac{v - u}{t}$

$a = \frac{14.7 - 0}{3.5}$

$a = 4.2m \: s ^{ - 2}$

4 0

2 years ago

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

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

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

2 years ago

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

Alisiya [41]

Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

7 0

2 years ago

A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s

Oksana_A [137]

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

5 0

2 years ago