Ask question

Ask question

All categories

erma4kov [3.2K]

3 years ago

5

Astone is thrown directly upward with an initial speed of 9.6 m/s from a height of 12.8 m. After what time interval (in s) does

the stone strike the ground? Use g 9.8 m/s^2 Enter a number with 2 digits behind the decimal point.

1 answer:

lara31 [8.8K]3 years ago

3 0

Answer:

1.89 seconds

Explanation:

t = Time taken

u = Initial velocity = 9.6 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=9.6-9.8\times t\\\Rightarrow \frac{-9.6}{-9.8}=t\\\Rightarrow t=0.97 \s$

Time taken to reach maximum height is 0.97 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=9.6\times 0.97+\frac{1}{2}\times -9.8\times 0.97^2\\\Rightarrow s=4.7\ m$

So, the stone would travel 4.7 m up

So, total height ball would fall is 4.7+12.8 = 17.5 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 17.5=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{17.5\times 2}{9.8}}\\\Rightarrow t=1.89\ s$

Time taken by the stone to travel 17.5 m is 1.89 seconds

You might be interested in

TRUE OR FALSE! PLZ HELP

Ksju [112]

Answer:

True

Explanation:

Magnitude is the "value" the greater the value the greater the force is and vice versa

5 0

2 years ago

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

natali 33 [55]

With acceleration

$\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j$

and initial velocity

$\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i$

the velocity at time <em>t</em> (b) is given by

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}$

$\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

We can get the position at time <em>t</em> (a) by integrating the velocity:

$\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du$

The particle starts at the origin, so $\mathbf x(0)=\mathbf0$ .

$\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du$

$\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}$

$\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j$

Get the coordinates at <em>t</em> = 8.00 s by evaluating $\mathbf x(t)$ at this time:

$\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j$

$\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j$

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating $\mathbf v(t)$ at the same time:

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j$

$\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

$\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}$

5 0

3 years ago

The pull is always toward the _____ of the planet

Nataly [62]

The correct answer is center. Hope this helps.

8 0

3 years ago

Read 2 more answers

A bowling ball weighing 71.2 N is attached to the ceiling by a 3.30 m rope. The ball is pulled to one side and released; it then

wolverine [178]

Answer:

a. The acceleration of the bowling ball is 9.5 m/s² toward the center

b. The tension in the rope is 140.24 N

Explanation:

given information:

ball weight, W = 71.2 N

the length of rope, R = 3.30 m

ball speed, v = 5.60 m/s

a. The acceleration of the bowling ball, α

α = $\frac{v^{2} }{R}$

where

α = the acceleration

v = the speed

R = radius

thus

α = $\frac{v^{2} }{R}$

= $\frac{5.60^{2} }{3.3}$

= 9.5 m/s² toward the center

b. The tension in the rope?

according to the Newton's second law

ΣF = m a

where

F = force

m = mass

a = acceleration

so,

ΣF = m a

T - W = m a

T = m a + W

= (W a/g) + W

= (71.2 x 9.5/9.8) + 71.2

= 140.24 N

4 0

3 years ago

An object in equilibrium has three forces exerted on it. a 42 n force acts at 90° from the x-axis and a 46 n force acts at 60° f

Olegator [25]

Since the object is in equilibrium, then the sum of the forces acting on it is zero. You must do the sum of forces for the component "y" and for the component "x". From there you clear the components of the third force. Then you must find the direction and magnitude.
The result is a force of 85.01N at 74.30 degrees from the x axis. Attached solution.

8 0

3 years ago