Posted by **Jerry** on Thursday, September 13, 2012 at 2:41pm.

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.)

- Physics -
**Scott**, Thursday, September 13, 2012 at 7:07pm
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

- Physics -
**Jerry**, Thursday, September 13, 2012 at 7:20pm
Im sorry but what is the equation to use t in? Im confused by your wording.

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