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

1. Nov 23, 2009

### 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.

2. Nov 23, 2009

### Jebus_Chris

What equations do you have?

3. Nov 23, 2009

### XanziBar

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?:

4. Nov 23, 2009

### Identity

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