Answer to Question #9818 in Mechanics | Relativity for Sarah-Eve Turcotte
It's about projectile.
Here is the question:
A projectile is launched from a height of 20 m with a velocity oriented 30 degrees above the horizontal. If the height reached is 45m, what is the initial velocity of the shot?
Height actually reached h = 45 - 20 = 25 m.
is vertical component of initial velocity, so Vy = V * sin(30) = 0.5 * V.
V^2 / 2 = m * g * delta(h). As horizontal component of velocity is
Vy ^ 2 = 2 * g * 25
Vy = sqrt (50 * g)
V = Vy / sin(30) = 2 *
sqrt (50 * g) = 44.295 m/s.
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