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3 years ago

A ball is kicked horizontally from a 60 meter tall cliff at 10 m/s. How far from

Physics

1 answer:

o-na [289]3 years ago

6 0

Hi there!

We can begin by deriving the equation for how long the ball takes to reach the bottom of the cliff.

$\large\boxed{\Delta d = v_it+ \frac{1}{2}at^2}}$

There is NO initial vertical velocity, so:

$\large\boxed{\Delta d= \frac{1}{2}at^2}}$

Rearrange to solve for time:

$2\Delta d = at^2\\\\t = \sqrt{\frac{2\Delta d}{g}}$

Plug in the given height and acceleration due to gravity (g ≈ 9.8 m/s²)

$t = \sqrt{\frac{2(60)}{(9.8)}} = 3.5 s$

Now, use the following for finding the HORIZONTAL distance using its horizontal velocity:

$\large\boxed{d_x = vt}\\\\d_x = 10(3.5) = \karge\boxed{35 m}$

You might be interested in

What happens to the forces acting on a falling object at terminal velocity?

xenn [34]

Answer: Objects falling through a fluid eventually reach terminal velocity. At terminal velocity, the object moves at a steady speed in a constant direction because the resultant force acting on it is zero.

8 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

The amount of space an object takes up is its

Likurg_2 [28]

Volume.

Hope this helps.

r3t40

4 0

3 years ago

When 23Na is bombarded with protons, the products are 20Ne and A. a neutron B. an alpha particle C. a deuteron D. a gamma ray pa

Vikki [24]

Answer:

The correct option is B. an alpha particle.

Explanation:

When ²³Na is bombarded with protons $_{1}^{1}\textrm{H}$ , ²⁰Ne and one alpha particle $_{2}^{4}\textrm{He}$ is released.

The reaction is as follows:

$_{11}^{23}\textrm{Na} + _{1}^{1}\textrm{H} \rightarrow _{10}^{20}\textrm{Ne} + _{2}^{4}\textrm{He}$

Therefore, an alpha particle and ²⁰Ne are released when ²³Na is bombarded with protons.

5 0

4 years ago

If a Ferrari, with an initial velocity of 10m/s, accelerates at at rate of 50m/s/s for 3 seconds, what will its final velocity b

seraphim [82]

Answer:

160 m/s

Explanation:

The Ferrari is moving by uniformly accelerated motion, with constant acceleration of a = 50 m/s^2, and initial velocity u = 10 m/s. The velocity at time t of the car is given by

$v(t)= u +at$

where

u = 10 m/s

a = 50 m/s^2

If we substitute t = 3 s into the equation, we can find the velocity of the car after 3 seconds:

$v(3 s)=10 m/s + (50 m/s^2)(3 s)=160 m/s$

5 0

3 years ago