# Projectile Motion velocity homework

• _physics_noob_
In summary, a projectile is fired horizontally from a gun 55.0 m above flat ground with a speed of 350 m/s. After calculating for various parameters, it can be determined that the projectile remains in the air for 3.35 seconds, strikes the ground at a horizontal distance of 1190 m, has a y-component velocity of -33.32 m/s, a magnitude of speed of 351.6 m/s, and an angle of -5 degrees when it hits the ground. However, it is more common to use the angle of 175 degrees as it is the smaller angle with the surface.
_physics_noob_

## Homework Statement

A projectile is fired horizontally from a gun that is 55.0 m above flat ground, emerging from the gun with a speed of 350 m/s.
a) How long does the projectile remain in the air?
b) At what horizontal distance from the firing point does it strike the ground?
c) What is the y-component of the velocity as it strikes the ground?
d) What is the magnitude of its speed as it strikes the ground?
e) At what angle does the projectile strike the ground?

## Homework Equations

V1x=V1cosθ
V2x=V1cosθ
x=(V1cosθ)t
V1y=V1sinθ
V2y=(V1ysinθ)t-gt
y=(V1sinθ)t-0.5gt^2

## The Attempt at a Solution

So I attempted each one and just want feed back on my answers.
a)t=3.4s
b)x=1190m
c)-33.32m/s
d)351.6m/s
e)355 degrees

1. The precision of the time result is rather low, so some of the significant digits in the subsequent results are incorrect.

2. 355 degrees cannot be an angle with which the projectile strikes the ground, because this is underground. The angle must be less than 90 degrees.

So the timing is incorrect?
This is what I did...
y=(350m/sin(0))t-0.5gt^2
-55m=-0.5(9.8m/s^2)t^2
-55m/-4.9m/s^2=t^2
t=3.35s so I rounded it to 3.4s

Last edited:
voko said:

1. The precision of the time result is rather low, so some of the significant digits in the subsequent results are incorrect.

2. 355 degrees cannot be an angle with which the projectile strikes the ground, because this is underground. The angle must be less than 90 degrees.

Not necessarily.

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Im not sure of you question. Adding 360 to 355 would get you to 355 again, you would make a complete circle. If I have interpreted your answer correctly, 355 into the ground, equals 175 above y axis. I have not worked the problem though.

3.35 s is better.

For an angle with a surface, one may may always choose either an angle greater than 90, or less than 90. Usually the smaller angle is selected.

Okay so I took the tanarc(-33.32/350) and got -5 degrees... Then I added 360 to get 355 degrees... So I would add 180 degrees instead of 355 because 355 degrees would be the angle it penetrates the ground? and 175 degrees would be the angle at which the projectile hit the ground?

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It involves both horizontal and vertical motion, and the object follows a curved path known as a parabola.

## 2. How is the velocity of a projectile calculated?

The velocity of a projectile can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time the object is in motion.

## 3. What factors affect the velocity of a projectile?

The velocity of a projectile is affected by the angle of launch, the initial velocity, air resistance, and the acceleration due to gravity. Other factors such as wind, temperature, and air density can also have an impact on the velocity.

## 4. How does the velocity of a projectile change during its flight?

The velocity of a projectile changes continuously during its flight due to the influence of gravity. Initially, the velocity decreases in the vertical direction due to the force of gravity, while the horizontal velocity remains constant. As the projectile reaches its highest point, the vertical velocity becomes zero and then starts to increase in the opposite direction. The horizontal velocity remains constant throughout the entire flight.

## 5. What is the relationship between the angle of launch and the velocity of a projectile?

The angle of launch has a direct impact on the velocity of a projectile. The greater the angle of launch, the greater the vertical velocity component and the longer the projectile stays in the air. However, if the angle of launch is too high, the horizontal velocity component decreases, resulting in a shorter horizontal distance traveled. The optimal angle of launch for maximum distance is 45 degrees.

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