Projectiles

BabyMilo

A particle of Q is projected from A at ground level. After 4 seconds the particle strikes the ground again at B where AB = 200m. Calculate the initial velocity of Q.

using A=theta
Equation 1
$$\displaystyle 2usinA-2g=0$$

$$\displaystyle u=\frac{2g}{2sinA}$$

Equation 2
$$\displaystyle 4ucosA=200$$

Sub 1 in 2

$$\displaystyle 4*\frac{2g}{2sinA}cosA=200$$

$$\displaystyle tanA=\frac{8g}{400}$$

then $$\displaystyle A=11.1$$
and $$\displaystyle u=50.95ms^{-1}$$

is everything correct?

Ackbeet

MHF Hall of Honor
Not sure where you're getting your equations from. The range of the projectile is given by

$$\displaystyle r=\frac{v_{i}^{2}}{g}\,\sin(2\theta)$$.

The time to impact is given by

$$\displaystyle t=\frac{2v_{iy}}{g}$$.

Finally, you're going to have that

$$\displaystyle v_{iy}=v_{i}\sin(\theta)$$.

Solve these equations for $$\displaystyle v_{i}$$.

Last edited:

BabyMilo

Not sure where you're getting your equations from. The range of the projectile is given by

$$\displaystyle r=\frac{v_{i}^{2}}{g}\,\sin(2\theta)$$.

The time to impact is given by

$$\displaystyle t=\frac{2v_{iy}}{g}$$.

Finally, you're going to have that

$$\displaystyle v_{iy}=v_{0}\sin(\theta)$$.

Solve these equations for $$\displaystyle v_{i}$$.
what do i do with the equations?
i don't really understand. the notations.

thanks

Ackbeet

MHF Hall of Honor
Ok. Here's an explanation of all symbols in these equations:

$$\displaystyle r$$ is the horizontal distance along the ground between the launch point and the striking point. Otherwise called the range.

$$\displaystyle v_{i}$$ is the magnitude of the initial velocity. Also called the initial speed of the projectile. This is a scalar quantity, and is equal to the magnitude of the initial velocity vector.

$$\displaystyle g$$ is the acceleration due to gravity.

$$\displaystyle \theta$$ is the angle of elevation of the projectile when launched. It's the angle between the horizontal and the initial velocity vector.

$$\displaystyle t$$ is the amount of time it takes the projectile to strike the ground after having been launched.

$$\displaystyle v_{iy}$$ is the component of the initial velocity in the $$\displaystyle y$$-direction.

There's a notational error in my equations:

$$\displaystyle v_{iy}=v_{i}\sin(\theta)$$, not $$\displaystyle v_{iy}=v_{0}\sin(\theta)$$. Sorry about that. I've fixed that. Does this help explain things?

Unknown008

If you want to work form the basic principles;

you know that

$$\displaystyle s = ut + \frac12 at^2$$

Let v be the initial velocity.

Taking along the vertical;

$$\displaystyle 0 = vsin\theta (4) - \frac12 9.8 (4)^2$$

$$\displaystyle 0 = 4vsin\theta - 78.4$$

$$\displaystyle v sin\theta = 19.6$$

Taking along the horizontal, there is no acceleration due to gravity:

$$\displaystyle 200 = vcos\theta (4) + \frac12 (0)(4)$$

$$\displaystyle 200 = 4vcos \theta$$

$$\displaystyle v cos\theta = 50$$

Solve for v by eliminating the angle of projection theta.