1. ## Projectiles

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
$2usinA-2g=0$

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

Equation 2
$4ucosA=200$

Sub 1 in 2

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

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

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

is everything correct?

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

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

The time to impact is given by

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

Finally, you're going to have that

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

Solve these equations for $v_{i}$.

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

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

The time to impact is given by

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

Finally, you're going to have that

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

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

thanks

4. Ok. Here's an explanation of all symbols in these equations:

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

$v_{i}$ 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.

$g$ is the acceleration due to gravity.

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

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

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

There's a notational error in my equations:

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

5. If you want to work form the basic principles;

you know that

$s = ut + \frac12 at^2$

Let v be the initial velocity.

Taking along the vertical;

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

$0 = 4vsin\theta - 78.4$

$v sin\theta = 19.6$

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

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

$200 = 4vcos \theta$

$v cos\theta = 50$

Solve for v by eliminating the angle of projection theta.