quadratics / projectiles.
take: h(t) = -4.905 t² + Viy t + hi
where h is height above ground in meters
t is time in seconds
Viy is the initial vertical speed in m/s
and hi is the initial height in meters.
if you want a height as a function of horizontal distance thingy [something much more useful to you] :
h(d) = (-4.905 / Vix²) d² + (Viy / Vix) d + hi
where:
h is height above ground in meters
d is horizontal distance in meters
Vix is initial horizontal velocity in m/s
Viy is initial vertical velocity in m/s
hi is height above the ground, in meters
now, for some trig:
if the person is hitting the volleyball at an angle, with a constant inital velocity, you hafta break the initial velocity into it's x and y parts:
Vix = Vi cos º
Viy = Vi sin º
where:
Vi is the initial velocity, at an angle in standard position
Vix is the initial horizontal velocity
Viy is the initial vertical velocity
º is the angle in degrees, in standard position.
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/|
/ |
Vi / |
/ | Viy
/ _|
/º__|_|
Vix
standard position:
ok, pretend the following addition sign is a graph, with all four quadrants
+
the prong on the right would be 0º in standard position, straight up would be 90º.
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90º
180º + 0º
270º
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