ITS EASY TRUST ME!!!!

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

5 0

3 years ago

How far will a free falling object fall in 8.7 secs if it started from rest? Remember acceleration is negative for free fall. Do

sleet_krkn [62]

Answer:

h~=371.26m

Explanation:

when an object falls we use the equations of accelerated motion. There is only one that gives distance.

$x = ut + \frac{1}{2} a {t}^{2}$

Since we have no initial velocity (started from rest) we can get rid of the (ut) term

where a we substitute g (gravitational acceleration, constant for given heights and almost 9.81m/s^2).

$h = \frac{1}{2} g {t}^{2} = \frac{1}{2} \times 9.81 \times {8.7}^{2} = 371.26m$

4 0

3 years ago

Someone please help me answer these

nekit [7.7K]

1. GPE

2. KE

3. KE

4. KE

5. Both

6. Both

7. Neither

8. Neither

Alright I think these should be right ;)

4 0

3 years ago

Which of the following is NOT a true statement about error in hypothesis testing?Choose the correct answer below.A.The symbol a

Marizza181 [45]

Answer:

Option D

A type I error is making the mistake of rejecting the null hypothesis when it is actually false.

Explanation:

Error type I is usually represented by alpha symbol and type I error entail making a mistake of rejecting the null hypothesis when it's actually true. Type II error on the other side involves making a mistake of failing to reject null hypothesis when it is actually false. The statement in option D is false because it contradicts the definition of type I error above hence the only false statement in relation to hypothesis testing is option D, A type I error is making the mistake of rejecting the null hypothesis when it is actually false.

8 0

3 years ago

Please help with physics 2 part question giving 50 points

AveGali [126]

Answer:

2.88×10⁻⁹ s

2.40×10¹⁵ m/s²

Explanation:

Given:

v₀ = 12300 m/s

v = 6.92×10⁶ m/s

Δx = 0.997 cm = 0.00997 m

Part 1) Find: t

Δx = ½ (v + v₀)t

0.00997 m = ½ (6.92×10⁶ m/s + 12300 m/s) t

t = 2.88×10⁻⁹ s

Part 2) Find: a

(6.92×10⁶ m/s)² = (12300 m/s)² + 2a (0.00997 m)

a = 2.40×10¹⁵ m/s²

5 0

3 years ago