Ask question

Ask question

All categories

3 years ago

15

A cannonball is fired horizontally from the top of a cliff. the cannon is at height h = 80.0 m above ground level, and the ball

is fired with initial horizontal speed v0. assume acceleration due to gravity to be g = 9.80 m/s2 .assume that the cannon is fired at time t=0 and that the cannonball hits the ground at time tg. what is the y position of the cannonball at the time tg/2?

1 answer:

SOVA2 [1]3 years ago

6 0

Answer;

= 60 m

Explanation;

s = 80 m

= ½at² = ½ * 9.8m/s² * t² → t = 4.04 s

At any time 0 ≤ t ≤ 4.04 s, height is

y = Yo + ½at² = 80 m - 4.9m/s² * t²

so at t = 4.04s/2 = 2.02 s,

y = 80 m - 4.9m/s² * (2.02s)² = 60 m

(so it drops 15 m during the first 2.02 s and 60 m during the second 2.02 s)

You might be interested in

An earthquake causes a 3 kg book to fall from a shelf. If the book lands with

Anna [14]

The correct answer is C

7 0

3 years ago

Narysuj wykres zależności szybkości od czasu i drogi od czasu jeśli ciało porusza się ruchem jednostajnym z szybkością 45 m/s.

murzikaleks [220]

Lett me come back imma translate this... and then ill come to help

7 0

3 years ago

Find the angle for the third-order maximum for 591 nm wavelength light falling on a diffraction grating having 1460 lines per ce

Marat540 [252]

Answer:

15.32°

Explanation:

We have given the wavelength $\lambda =591nm=591\times 10^{-9}m$

Diffraction grating is 1460 lines per cm

So $d=\frac{10^{-2}}{1460}=6.71\times 10^{-6}m$ (as 1 m=100 cm )

For maximum diffraction

$dsin\Theta =m\lambda$ here m is order of diffraction

So $6.71\times 10^{-6}sin\Theta =3\times 591\times 10^{-9}$

$sin\Theta =0.264$

$\Theta =15.32^{\circ}$

6 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

What is a gas-like mixture that is made of charged particles?<br><br> (Physical Science)

Juli2301 [7.4K]

Answer:

I believe the answer is Plasma

7 0

3 years ago