Posted by **Amy** on Wednesday, January 18, 2012 at 3:40pm.

A small ball is launched at an angle of 30.0 degrees about the horizontal. It reaches a maximum height of 2.5m with a respect to the launch position.

What is the initial velocity of the ball when its launched?

What is the range, defined as a horizontal distance traveled until it returns to his original height. (you can ignore air resistance)

- Physics -
**Henry**, Friday, January 20, 2012 at 3:23pm
Yf^2 = Yo^2 + 2g*h,

Yo^2 = Yf^2 - 2g*h,

Yo^2 = o - (-19.6)*2.5 = 960.4,

Yo = 31 m/s = Ver. component of Vo.

a. Vo = Yo/sinA = 31/sin30 = 62 m/s = Initial velocity.

b. R = Vo^2*sin(2A)/g,

R = (62)^2*sin60/9.8 = 339.7 m.

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