help
i just wanted to ask you first, have you learned about projectiles already?
na not really.. this was a blast. my prof didn't discussed this to us
because if you learned that, i will know that we need to include the acceleration due to gravity since it's dealing with vertical motions. if you did not learned that yet, maybe it's only dealing with one dimension.if this is the case we will just use the velocity formula \(v=\frac{\Delta d}{\Delta t}\). but anyways, i know it's a projectile so i think we need to use the five kinematics equations, are you familiar with that?
its 4 i think? yes.. d= (Vinitial +Vfinal/T)*t
Vfinal/ Vinitial + at, Vf^2= Vi^2 + 2ad, d= Vinitial(time) + 1/2 at^2
aah, that's good then :) okay let's start by drawing the diagram, can you do that for me?
diagram? im so sorry but i have no idea..
just draw the situation from the problem, like the stone falling
great:) so next, we put our givens: dy=200 m ay=9.8m/s^2 v=5m/s --> not sure if this is the initial velocity, what do you think? let's assume that it is the initial velocity t=? to find the time we use: \(d_y=v_it + \frac{1}{2} a_y t^2\) substitute the givens and you will end up having a quadratic equation, so you need to use the quadratic formula to solve for "t" note: i use subscript "y" to represent vertical component
okay, ill solve for t.. ill try. hehe
okie dokie :)
t1= 5.90 and t2= -6.92
hmmm.. i got a different answer? O.o
\(200= 5t + 4.9t^2\) \( 0=4.9t^2+5t-200\) i set [down] as positive and [up] is negative
i guess, i'll take the negative?
if you will set down as negative it should be the same \(-200=5t -4.9t^2\) \(4.9t^2 -5t -200=0\)
oops i think i messed up somewhere.. sorry.. lemme try it again
its -5.90 and 6.92
yeah you're right XD
so the time would be??
=)
there's no negative time, so you should take the positive
ah common sense silver tsk tsk haha thank you!!!!!
np:)
hmm.. not sure, it's either the height of the tower, or the velocity of the second object
ah yes.. yes well. hehe THANK YOU SO MUCH =)
(^_^)
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