sensei's last words - Unit 4

 Projectile motion gets even more complicated than the vegas rule, and 2 dimensional analysis when angles are added to the word problem. But not to worry, if you know your trigonometry than it's quite easy to understand.

Here is a problem I made up, that involves angles in projectile motion:
   
          If a girl jumps off a rock ledge 5 meters high into a pool with a velocity of 2m/sec at an angle of 5˚, what is the initial horizontal and vertical velocity, how long is the girl in the air, and how far horizontally will the girl land in the pool?

First you have to do the following trigonometry to get the initial horizontal and vertical velocity.








Once you know the horizontal velocity, you can plug that in the XY table as the initial and final velocity on the x-axis. The vertical velocity can be plugged in as the initial velocity on the y axis. Next, we have to remember our rules from yesterday, and find the time in the aYer. That can be done with the "DAT" equation: Dy(m)=1/2 ay (m/s^2) t^2 (s) + Voy (m/s) t (s)
or
5m=1/2(-9.8m/s^2)t^2(s) + 0.17 (m/s) t
5=-4.9t^2+0.17t
4.9t^2-0.17t+5=0
(do quadratic formula)
t=1.03

Once you know the time, find the distance of the x-axis using the DAT equation again.

D=1/2(0)1.03^2+1.99(1.03)
D=2.58
She went 2.58 meters horizontally into the pool.

Summary statement, a girl jumps off a rock ledge 5 meters high into a pool with a velocity of 2m/sec at an angle of 5˚, what is the initial horizontal velocity of 1.99 m/s and an initial vertical velocity of 0.17m/s. The girl was in the air 1.03 seconds, and jumped a horizontal distance of  2.58 meters.