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.

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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