# How Do You Solve a Projectile Motion Problem?

• blackcemre
In summary: One for before firing and one for after firing. Can you please clarify which equation is for before firing?B.)Determine the total time t in the air.t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 sec
blackcemre

## Homework Statement

A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.

1-Determine the maximum height reached by the projectile.

2-Determine the total time in the air.

3-Determine the total horizontal distance covered (that is, the range).

4-Determine the speed of the projectile 1.50 s after firing.

5-Determine the direction of the motion of the projectile 1.50 s after firing.

## Homework Equations

v = vo +at x = xo + vot + .5at2 v2 = vo2 + 2a(x - xo) x=vt

## The Attempt at a Solution

A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 mB.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 secC.)Determine the total horizontal distance covered (that is, the range).

Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m

Last edited by a moderator:
blackcemre said:

## Homework Statement

A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.

1-Determine the maximum height reached by the projectile.

2-Determine the total time in the air.

3-Determine the total horizontal distance covered (that is, the range).

4-Determine the speed of the projectile 1.50 s after firing.

5-Determine the direction of the motion of the projectile 1.50 s after firing.

## Homework Equations

v = vo +at x = xo + vot + .5at2 v2 = vo2 + 2a(x - xo) x=vt

## The Attempt at a Solution

A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 mB.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 secC.)Determine the total horizontal distance covered (that is, the range).

Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m

So what's the question? You couldn't do #4 and #5?

Zz.

blackcemre said:
A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal

blackcemre said:
A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 m
The numbers you've used in your work do not match those stated in the problem statement. Which numbers are correct?

There is confusion about the symbols used.
Is V the initial speed?
In the relevan equations, there are two different equations for x.

## 1. What is projectile motion?

Projectile motion is the motion of an object that is launched or thrown into the air and moves along a curved path under the influence of gravity.

## 2. What factors affect the trajectory of a projectile?

The trajectory of a projectile is affected by the initial velocity, the angle of launch, and the force of gravity.

## 3. How is the range of a projectile calculated?

The range of a projectile is calculated using the formula R = (V^2 * sin(2θ))/g, where V is the initial velocity, θ is the angle of launch, and g is the acceleration due to gravity.

## 4. Can a projectile have a curved trajectory?

Yes, a projectile can have a curved trajectory if there is an external force acting on it, such as air resistance or wind.

## 5. What is the maximum height reached by a projectile?

The maximum height reached by a projectile is calculated using the formula H = (V^2 * sin^2(θ))/(2g), where V is the initial velocity and θ is the angle of launch. This is the point at which the vertical component of the projectile's velocity becomes zero.

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