Ask question

Ask question

All categories

nataly862011 [7]

3 years ago

15

A rock is thrown from the top of a 20-m building at an angle of 53Â° above the horizontal. If the horizontal range of the throw

is equal to the height of the building, with what speed was the rock thrown? What is the velocity of the rock just before it strikes the ground?

1 answer:

NemiM [27]3 years ago

6 0

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

v = 18.22 m/s

You might be interested in

9. How does the length of the hypotenuse in a right triangle relate to the lengths of the legs? (2 points)

konstantin123 [22]

<span>The pythagorean theorem addresses the length of the hypotenuse in relation to the length of the legs. The square root of the length of the hypotenuse is equal to the sum of one leg squared plus the other leg squared. In other words, A squared plus B squared equals C squared where A and B are the lengths of the legs of the triangle and C is the length of the hypotenuse.</span>

6 0

3 years ago

Read 2 more answers

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

It takes Mars about ____ Earth years to orbit the Sun.

OlgaM077 [116]

<span>At this distance, and with an orbital speed of 24.077 km/s, Mars takes 686.971 Earth days, the equivalent of 1.88 Earth years, to complete a orbit around the Sun. This eccentricity is one of the most pronounced in the Solar System, with only Mercury having a greater one (0.205).

686.971 rounds to 687
HOPE I HELPED!</span>

7 0

3 years ago

Read 2 more answers

Two metals of equal mass with different heat capacities are subjected to the same amount of heat. which undergoes the smallest c

liraira [26]

<span>So when two metals of equal mass but different heat capabilities are subjected to same heat quantity, the metal with higher heat capacity have the small temperature change. Heat supplied is determined as heat capacity of the metal times the change in temperature.</span>

8 0

4 years ago

Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the

Assoli18 [71]

Answer:

Part a)

$\rho = 3\epsilon_0 k r^2$

Part b)

$Q = 4\pi \epsilon_0kR^5$

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

$E = kr^3 \hat r$

now electric flux through a spherical surface of radius r is given as

$\phi = E. A$

$\phi = kr^3(4\pi r^2)$

now by Guass law we know that

$E.A = \frac{q}{\epsilon_0}$

$q = 4\pi \epsilon_0kr^5$

now volume charge density is given as

$\rho = \frac{q}{\frac{4}{3}\pi r^3}$

$\rho = 3\epsilon_0 k r^2$

Part b)

Total charge inside the radius R is given as

$Q = 4\pi \epsilon_0kR^5$

7 0

3 years ago