All categories

3 years ago

Which best illustrates projectile motion

Physics

2 answers:

Gwar [14]3 years ago

6 0

Answer:

its D

Explanation:

took the test

worty [1.4K]3 years ago

3 0

Answer:

D) The guy jumping

You might be interested in

Workers do 8000 J of work on a 2000-N crate to push it up a ramp. If the ramp is 2 m high, what is the efficiency of the ramp?

IRISSAK [1]

Answer:

50%

Explanation:

Efficiency = work out / work in

e = Fd / W

e = (2000 N) (2 m) / (8000 J)

e = 0.5

7 0

4 years ago

Read 2 more answers

is the light transmitted by a filter brighter than, less brighter than, or the same brightness as the light that hits it? explai

vichka [17]

The light transmitted by a filter is less bright than the light that hits it.

A filter only absorbs/removes part of the light that hits it. That means it removes some energy from the incident light. The remaining light carries less energy, so it must be less bright.

5 0

4 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: