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Physics, Projectile Motion Help?
Ive been absent from school for the entire week and I don't know how to do these.
http://i.imgur.com/GTcEMrI.png
Could you help me do some of these? or explain how to do some? This is due in 45 mins. Ive spent past 2 hours trying to figure out how to do this and this is so frustrating!
Any question that you help on would be highly appreciated.
Thanks!
1 Answer
- GeronimoLv 77 years agoFavorite Answer
Vix = (29.8) • cos(24º) ... horizontal component of the initial velocity
Viy = (29.8) • sin(24º) ... vertical component of the initial velocity
Since air resistance (friction) is neglected, the horizontal component of the
initial velocity is CONSTANT: Vix = Vx
The horizontal displacement given is: ∆x = 67.2724 m
so the time of flight ( T ) is found using the horizontal motion equation:
∆x = (Vix) • T ... distance = speed • time
67.2724 = (29.8) • cos(24º) • T
T = 2.47 sec ... time of flight ◀◀
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
At the peak height, the vertical component of the velocity is ZERO.
Using the standard energy-motion physics equation:
(Vғy)² – (Viy)² = 2 • g • h ... final vertical velocity = Vғy and height = h
(0)² – ((29.8) • sin(24º))² = 2 • (- 9.81) • h ... where: Vғy = 0 at the peak height
h = 7.49 meters ... peak height
