Standard temperature and pressure are defined as

A ball is thrown straight up. It passes a 2.00-m-high window 7.50 m off the ground on its path up and takes 0.312 s to go past t

marta [7]

Answer:

Explanation:

Given that

The window height is 2m

And the window is 7.5m from the ground

Then the total height of the window from the ground is 7.5+2=9.5m

It takes the ball 0.32sec travelled pass the window.

When the ball get to the window, it has an initial velocity (u') and when it gets to the top of the window it has a final velocity ( v')

Now using the equation of free fall during this window travels

S=ut-½gt² against motion.

S=2, g=9.81, t=0.32sec

Then,

S=u't-½gt²

2=u'×0.32-½×9.81×0.32²

2=0.32u'-0.5023

2+0.5032=0.32u'

Then, 0.32u'=2.5032

u'=2.5032/0.32

u'=7.82m/s

This is the initial velocity as the ball got the the window

Now, let analyse from the window bottom to the ground which is a distance of 7.5m

Using the equation of free fall again

v²=u²-2gH

In this case the final velocity (v) is the velocity when the ball reach the bottom of the window i.e u'=7.82m/s,

While u is the original initial velocity from the throw of the ball

Then,

u'²=u²-2gH

7.82²=u²-2×9.81×7.5

61.146=u²-147.15

61.146+147.15=u²

Then, u²=208.296

So, u=√208.296

u=14.43m/s

The initial velocity of the ball form the throw is 14.43m/s

6 0

3 years ago

What is nuclear energy

suter [353]

Answer:

Nuclear energy, also called atomic energy, energy that is released in significant amounts in processes that affect atomic nuclei, the dense cores of atoms. It is distinct from the energy of other atomic phenomena such as ordinary chemical reactions, which involve only the orbital electrons of atoms.

Explanation:

7 0

3 years ago

Read 2 more answers

Tom rides his motorcycle at a speed of 15 meters/second for an hour.

Dahasolnce [82]

Answer:

1 hour to ride his motorcycle

4 0

3 years ago

Which reading strategy helps you to read and understand a passage quickly by making use of your previous knowledge?

Molodets [167]

Taking notes,because you can read back on what you learned

4 0

3 years ago

A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.

Ne4ueva [31]

The range of a projectile motion is given by:

$\frac{u_o^2 sin2\theta}{g}$

where, $u$ is the initial speed of the projectile, $\theta$ is the angle of the projectile and $g$ is the acceleration due to gravity.

The maximum height reached is given by:

$\frac{u_o^2 sin^2\theta}{2g}$

Part a

It is given that the maximum height reached is equal to the horizontal range. we need to find the angle of the projectile.

Equating the two:

$\frac{u_o^2 sin2\theta}{g}=\frac{u_o^2 sin^2\theta}{2g}\\ \Rightarrow 2 sin2\theta =sin^2\theta\\ \Rightarrow 2\times 2 sin\theta cos\theta=sin^2 \theta\\ \Rightarrow tan\theta =4\\ \Rightarrow \theta=tan^{-1}4=75.96^o$

Hence, the projectile was thrown at an initial angle of $\theta=75.96^o$ .

Part b

we need to find the angle for which range would be maximum and then write this maximum range in terms of original range.

So, we know that range is given by:

$R= \frac{u_o^2sin2\theta}{g}$

It would be maximum when $sin2\theta=1\Rightarrow 2\theta=90^o\Rightarrow \theta=45^o$

Hence, $R_{max}=\frac{u_o^2}{g}$

Original range,

$R=\frac{u_o^2sin2\times75.96^o}{g}\\ \Rightarrow R=\frac{u_o^2}{g}\times 0.47\\ \Rightarrow R=R_{max}0.47\\ \Rightarrow R_{max}=2.125R$

Part c:

In the part a, we know that the angle of the projectile is independent of the $g$ i.e. the acceleration due to gravity and this is the only factor that varies with the different planets. Hence, the answer would remain same.

7 0

3 years ago