# Thread: Projectile motion and Vectors

1. ## 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.

2. 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

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

$\displaystyle v=29.15 m/s$

3. Thanks

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