An object is thrown upwards and after 2 seconds is 78 m above where it was initially thrown. Assume no air resistance, and determine the initial speed the object was thrown at.
Wrong group.
right group ... kinda
Sorta. Physics.
we are in meters so -4.9 assume 0 for initial height
\[s(t) = - \frac {1}{2} g t^2 + v_0 t + s_0\]
-4.6 t^2 + V t = 78
6, 9, not much difference eh
Just use the Kinematics equations, right?
what does 78m stand for? displacement, distance, or velocity?
kinetics are for babies and old ladies ....
Kinematics aren't kinetics. But okay.
78m has no direction, and also speed is m/s.
but it says upwards :s
If it says upwards, then its displacement.
|dw:1329871307054:dw| it might be easier to assume its a vertex
a minimum velocity perhaps :)
kk. thanks what is -4.9? how did we get that?
-4.9(4) + V(2) - 78 = 0 V = 39 +4.9(2) maybe -4.9 is gravity holding us back
since we are in meters
Wait. I'm stupid. Isn't gravity 9.8?
yeah gravity is 9.8.
acc due to g is 9.8 velocity = 9.8t position = 9.8t^2/2 = 4.9 t^2 but since gravity is a downward force; -4.9 t^2
Makes sense.
With that explained, I think we are done.
but .... im too old to die!!
"You're never to young to die"-A quote that makes me sad.
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