Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

A tennis ball is thrown vertically upwards with an initial velo. of 5m/s.Assuming the acceleration due to gravity is 10m/s^2. find the max gain in height of the ball.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- JamesJ

When the ball is at its maximum height all of its energy is in gravitational potential energy, PE. We can express the PE of an object of mass m at a height as
PE = mgh.
When it thrown, all of its energy is kinetic energy, KE. What's the expression for KE? When you know it, set it equal to the expression for PE and you'll be able to solve for h, the height.

- anonymous

but this is supposedly a kinematics question....

- JamesJ

This is kinematics

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- JamesJ

\[ KE = \frac{1}{2}mv^2 \]
Set KE = PE and you can find an expression for h ...

- anonymous

25/20meters...in ideal conditions

- JamesJ

If PE = KE then
\[ mgh = \frac{1}{2}mv^2 \]
Cancel m from both sides and divide by g, then
\[ h = \frac{v^2}{2g} \]
See now how to replicate antrio's answer?

- JamesJ

(@antrio: welcome to OpenStudy! fyi, we try and avoid giving answers outright. Much better to show people how to get to the answer themselves.)

- anonymous

wong@ the questions can be solved by 3rd eqn of motion...or by conservation of energy.both yeilds same result....james is right

- JamesJ

Jamelynn, talk to me ... where are you stuck?

- anonymous

okay james@

- anonymous

i found out from my notes another way to solve the prob by using the equation for constant acceleration:
s=u^2 + 2as
s=5^2 + 2(-10)s
21s=25
s=1.19

- JamesJ

The equation you've written down is not correct.
What is correct is that if u is initial velocity and v final, then
\[ v^2 = u^2 + 2as \]
At the top of its arc, the object has no velocity, so v = 0. Hence
\[ s = - \frac{u^2}{2a} \]
The acceleration here is gravity, hence a = -g. Thus
\[ s = \frac{u^2}{2g} \]
This is identical to the equation above, become there v is also the initial velocity.
In short, your current answer is wrong.

- anonymous

there is mistake in that equ....the correct eqn is s=u x time + 1/2 g time^2..........and it cannot be applied here as time is not given

- anonymous

thx, i realized there was a mistake in my formula

- JamesJ

So one more time
All the kinetic energy when the ball is thrown is converted at the top of its flight into gravitational potential energy
KE(when ball thrown) = PE(when ball at the top of its flight)
Let \( u \) be the velocity at which it is thrown. Then
\[ KE = \frac{1}{2}mu^2 \]
Let \( s \) be the height at the top of its path. Then
\[ PE = mgs \]
Hence \( KE = PE \) implies
\[ \frac{1}{2}mu^2 = 2gs \]
Solving for s, we find that
\[ s = \frac{u^2}{2g} \]
Given that
\[ u = 5 \ m/s \]
and
\[ g = 10 \ m/s^2 \]
then
\[ s = \frac{u^2}{2g} = \frac{5^2}{2(10)} = \frac{25}{20} = 1.25 \ m \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.