Posted by Genevieve on Friday, September 17, 2010 at 4:19pm.
147= 0+ vxi(4.30407) + 4.9(4.30407)^2
well, I have issues with this. Why is gravity in the horizontal equation? (Besides,if it were, it would be negative).
In the yf=vyt+.5gt^2 , the last term should be negative (-.5gt^2). Upwards is postive, downward negative.
let V = initial velocity
xf =xi + vxi t
There is no acceleration in the horizontal direction, horizontal speed is constant!
vxi = V cos 37
so xf = 0 + .8 V t = 147
solve for V t
Now do y direction (up)
yf = yi + V t sin 37 - .5 (9.8) t^2
Note g is - if up is +
27 = 1 + V t sin 37 -4.9 t^2
we know V t
solve for t
and, on the yf, dont forget the initial height.
Yf=yi+Vy*t-1/2 g t^2
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