Physics
posted by jim .
Suppose the stone is thrown at an angle of 32.0° below the horizontal from the same building (h = 49.0 m) as in the example above. If it strikes the ground 69.3 m away, find the following. (Hint: For part (a), use the equation for the xdisplacement to eliminate v0t from the equation for the ydisplacement.)
(a) the time of flight s
(b) the initial speed m/s
(c) the speed and angle of the velocity vector with respect to the horizontal at impact
speed m/s
angle ° below the horizontal

heres two answers i got for the same question.
physics  bobpursley, Friday, May 20, 2016 at 4:49pm
break the initial velocity into two componnets;
horizontal: Vo*cos40
vertical: Vo*sin40  9.8 t
a. hf=42=vo*sin40*t1/2 9.8 t^2
d=36.7=vo*cos40t
solve for vo*t in the second in terms of t. Put that in the first equaiton.
b. knowing t, use the second equation to solve for vo.
physics  Damon, Friday, May 20, 2016 at 4:55pm
42 = Vo t + 4.9 t^2
Vo = speed sin 40
36.7 = u t
u = speed cos 40
so
speed * t = 36.7/cos 40
so
speed * sin 40 * t = 36.7 tan 40
Vo t = 36.7 tan 40
so back to
42 = Vo t + 4.9 t^2
now
42 = 36.7 tan 40 + 4.9 t^2
solve for t, time in the air, and go back and get the rest
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