Projectile motion and Vectors

• Aug 2nd 2009, 02:26 AM
DJ Hobo
Projectile motion and Vectors
I couldn't find a "Dynamics" section in the forum, so if this is in the wrong section, sorry for that.

At the moment, I'm studying Projectile motion using vectors and integration.

The question I'm stuck on is this:

A baseball player hits a ball to the outfield 50m away. The ball was in the air for 1.8 seconds. At what angle to the horizontal was the ball hit? At was speed was it hit?

Any help here would be greatly appreciated.

• Aug 2nd 2009, 03:20 AM
alexmahone
Quote:

Originally Posted by DJ Hobo
I couldn't find a "Dynamics" section in the forum, so if this is in the wrong section, sorry for that.

At the moment, I'm studying Projectile motion using vectors and integration.

The question I'm stuck on is this:

A baseball player hits a ball to the outfield 50m away. The ball was in the air for 1.8 seconds. At what angle to the horizontal was the ball hit? At was speed was it hit?

Any help here would be greatly appreciated.

Let the angle be $\displaystyle \theta$.

Let the speed be $\displaystyle v$.

Vertical motion:

t=1.8 s

Range=50 m

Horizontal motion:

$\displaystyle vcos\theta * t=50$

$\displaystyle vcos\theta*1.8=50$

$\displaystyle vcos\theta=\frac{500}{18}=\frac{250}{9}$ -------------- (1)

Vertical motion:

$\displaystyle s=vsin\theta\ t-\frac{1}{2}gt^2$

$\displaystyle 0=vsin\theta\ t-\frac{1}{2}gt^2$

$\displaystyle 0=vsin\theta-\frac{1}{2}gt$

$\displaystyle 0=vsin\theta-\frac{1}{2}9.8*1.8$

$\displaystyle vsin\theta=8.82$ ---------------------(2)

Dividing (2) by (1) we get

$\displaystyle tan \theta=0.31752$

$\displaystyle \theta=17.61^o$

Using this in (1), we get

$\displaystyle v*0.953=\frac{250}{9}$

$\displaystyle v=29.15 m/s$
• Aug 2nd 2009, 03:53 AM
DJ Hobo
Thanks :D

EDIT - I've tried the problem on my own now, and I've managed to do it.