Answer to Question #4581 in Mechanics | Relativity for Marcus
Let V1 and V2 be the vertical and the horizontal speed components. As the angle is pi/4, we see that V1 = V2 = V/sqrt(2).
Horizontal: distance D = V2*t = V*t/sqrt(2). (*)
Vertical: V1*t = (gt^2)/2, so, V*t/sqrt(2) = (gt^2)/2, and we see that D = (gt^2)/2. So, t = sqrt(2D/g) = sqrt(2*22.63/9.8) = 2.14 seconds.
From (*): the initial speed V = D*sqrt(2)/t = 22.63*sqrt(2)/2.14 = 14.95 m/s.
Let's find the maximum height.
Vt = (gt^2)/2, so, V = gt/2 and t = 2V/g = 3.05 seconds.
V - gt = 0 (0 is the speed at the top point), so time to get to the top point is t = V/g = 14.95/9.8 = 1.53 seconds.
Height H = Vt - (gt^2)/2 = 14.95*1.53 - (9.8*1.53^2)/2 = 11.40 meters.
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