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Answer to Question #9818 in Mechanics | Relativity for Sarah-Eve Turcotte

Question #9818
Hi I have a question for my physics' homework!
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?

Thank you
Expert's answer
Let initial velocity is V.
Height actually reached h = 45 - 20 = 25 m.
Vy
is vertical component of initial velocity, so Vy = V * sin(30) = 0.5 * V.
m *
V^2 / 2 = m * g * delta(h). As horizontal component of velocity is
constant,
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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