posted by Jerry .
From the edge of the rooftop of a building, a boy throws a stone at an angle 20.0° above the horizontal. The stone hits the ground 4.60 s later, 104 m away from the base of the building. (Ignore air resistance.)
What is the maximum height H reached by the stone?
(I found that the velocity is 24.06 and the initial height is 65.834, but I cant get the maximum height right.)
horizontal velocity (Vh) is the velocity (V) times the cosine of 20.0º
vertical velocity (Vv) is V times the sine of 20.0º
or __ Vv = Vh * tan(20.0º) = (104/4.60) * tan(20.0º) = 8.23
when the stone hits the ground, H = 0
0 = (-.5 * g * 4.60^2) + (8.23 * 4.60) + Ho
65.826 = Ho
use the axis of symmetry equation to find t at max H (x = -b / 2a)
t max = -8.23 / -9.8 = 0.84
substitute the t back to find H max
Im sorry but what is the equation to use t in? Im confused by your wording.