PHYSICS MIDDLE SCHOOL

019 10.0 pointsA ball is thrown horizontally from the top of
a building 120 m high. The ball strikes the
ground 64 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
020 (ment 1
210,

Answers

Answer 1

Answer:

Answer:

50.2 m/s

Explanation:

First of all, we need to find the time it takes for the ball to reach the ground.

The vertical position of the ball at time t is given by

$y(t) = h +ut + (1)/(2)gt^2$

where

h = 120 m is the initial height

u = 0 is the initial vertical velocity

g = -9.8 m/s^2 is the acceleration of gravity

The ball reaches the ground when y = 0. Substituting into the equation and solving for t, we find the time of light:

$0=h+(1)/(2)gt^2\nt=\sqrt{-(2h)/(g)}=\sqrt{-(2(120))/(-9.8)}=4.95 s$

The vertical component of the velocity of the ball changes following the equation

$v_y(t) = u+gt$

Substituting t = 4.95 s, we find the final vertical velocity of the ball just before reaching the ground:

$v_y=0+(-9.8)(4.95)=-48.5 m/s$

where the negative sign means the direction is downward.

We also can find the horizontal component of the velocity: since we know the horizontal distance travelled is d = 64 m,

$v_x = (d)/(t)=(64)/(4.95)=12.9 m/s$

And the final speed is calculated as the magnitude of the resultant of the two components of the velocity:

$v=√(v_x^2+v_y^2)=√(12.9^2+(-48.5)^2)=50.2 m/s$

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How is Ohm's Law related to house fires?

Answers

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a person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75 m ?

Answers

Answer:

Approximately $2.18 m\cdot s^(-1)$ .

Explanation:

Consider one of the equations for constant acceleration ("SUVAT" equations)

$v^(2) - u^(2) = 2 \; a \cdot x$ ,

where

$v$ is the final velocity of the object,
$u$ is the initial velocity of the object,
$a$ is the acceleration of the object, and
$x$ is the distance that the object had traveled while its velocity changed from $u$ to $v$ .

Note that unlike other SUVAT equations, this one does not ask for the time required for the speed of the object to change from $u$ to $v$ . Since in this problem, time isn't given, this time-less equation would particular useful.

Here

$v$ the final velocity needs to be found.
$u = 0$ for the stroller started from rest.
$a =\rm 0.500 \;m \cdot s^(-2)$ is the acceleration of the stroller, and
$x = \rm 4.75\; m$ is the distance that the stroller traveled while its velocity changed from $u$ to $v$ .

Rearrange the equation to isolate the unknown, $v$ :

$v^(2) = u^(2) + 2 \; a \cdot x$ .

Make sure that all units are standard, so that the unit of the output will also be standard. Apply the equation:

$v = \sqrt{u^(2) + 2 \; a \cdot x} = √(0^2 + 2 * 0.500 * 4.75 )\approx \rm 2.18\; m\cdot s^(-1)$ .

Hence the final velocity will be approximately $\rm 2.18 m\cdot s^(-1)$ .

A teacher points to a microscope sitting on the desk. Which conclusion best explainswhy the microscope is not moving?

Answers

Answer:

The object is sitting

Explanation:

Therefore the object cannot be in current motion as no force has been acted upon it (exept gravitational force)

Which trasformaron of energy takes place when a slingshot launches a stone ?

Answers

transformation of energy that takes place when a slingshot launches a stone is elastic potential energy tokinetic energy.

Will a cart rolling up and down a ramp have a faster or slower acceleration than a ball being thrown straight up into the air? (Assume up the ramp is positive and the origin is the starting point of the cart)

Answers

a ball being thrown straight up would excel faster.

Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart. What is the force applied between q1 and q2?