# Projectile motion problem - velocity , angle, and vertical acceleration unknown

• EL ALEM
In summary, the given information includes a projectile fired on a foreign planet with maximum horizontal distance of 40.2 m and maximum vertical distance of 25 m, reached in 25 seconds and 12.5 seconds respectively. The problem requires finding the initial velocity, angle at which it was fired, and acceleration due to gravity. Using equations for vertical and horizontal motion, the three unknowns can be solved for.
EL ALEM
This is all the given info:
On a foreign planet (g cannot equal 9.8m/s), a profectile is fired

Maximum horizontal distance= 40.2 m and it reaches this distance at 25s
Maximum vertical distance= 25 m and it reaches this distance at 12.5s

There is no horizontal acceleration, and there is no mass given (and I'm pretty sure its not needed to find the

solution)

What is the initial velocity of the projectile and angle at which it was fired and what is the acceleration due to

gravity

So basically I need to find vertical acceleration, intitial velocity, and angle theta.

I'm really stumped on this and all my attempts at a solution was just me listing out the givens.

What equations do you have?

So let's see... If this thing lands at the same vertical height it was launched at then (Vsin theta)^2/(2*y_max)="g" and the horizontal distance is Vi^2*sin(2*theta)/"g" and V*cos(theta)=distance in x/time. 3 equations, 3 unknowns.

Anyone want to look at this problem?:

Let the initial speed be v, the angle $$\theta$$ and acceleration due to gravity a.

For the vertical rise, use $$x = (v\sin{\theta})t+\frac{1}{2}at^2$$, where $$v\sin{\theta}$$ is the vertical velocity.

$$25 = v\sin{\theta}(12.5)+\frac{1}{2}a(12.5)^2$$

$$\Rightarrow 25 = 12.5 v\sin{\theta}+78.125a$$ --------(1)

Since the time it takes to fall back down is 25 seconds, we also have

$$0 = v\sin{\theta}(25)+\frac{1}{2}a(25)^2$$

$$\Rightarrow 0 = 25v\sin{\theta}+312.5a$$ --------(2)

For the horizontal distance, use $$x = (v\cos{\theta})t$$ (a = 0 in the horizontal component)

$$40.2 = v\cos{\theta}(25)$$ --------(3)

From equations (1) and (2), we can eliminate $$a$$ and $$v\sin{\theta}$$ (or just solve for a)

$$v\sin{\theta} = 4$$ --------------(4)

From equation (3),

$$v\cos{\theta} = 1.608$$ ----------(3')

To find $$\theta$$, divide (4) by (3')

To find $$v$$, square (4) and (3'), add them together, and use a trigonometric identity to simplify

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air or space under the influence of gravity. It follows a curved path called 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, and t is the time. Alternatively, it can also be calculated using trigonometry and the horizontal and vertical components of the initial velocity.

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

The range of a projectile is affected by the initial velocity, the angle of launch, and the vertical acceleration due to gravity. Air resistance and wind can also have an impact on the range.

## 4. How do you find the angle of launch for maximum range?

The angle of launch for maximum range can be found using the formula tanθ = (vy/vx), where θ is the angle of launch, vy is the vertical component of the initial velocity, and vx is the horizontal component of the initial velocity. The maximum range occurs when the angle of launch is 45 degrees.

## 5. What is the relationship between vertical and horizontal acceleration in projectile motion?

In projectile motion, the horizontal acceleration is constant and equal to zero, while the vertical acceleration is the acceleration due to gravity. This means that the horizontal velocity remains constant, while the vertical velocity changes due to the acceleration of gravity.

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