Projectiles

Feb 2008
383
38
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?

thanks for your help!
 

Ackbeet

MHF Hall of Honor
Jun 2010
6,318
2,433
CT, USA
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:
Feb 2008
383
38
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
Jun 2010
6,318
2,433
CT, USA
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?
 
May 2010
1,260
410
Mauritius
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.