Question

AN OBJECT IS LAUCHED AT A VELOCITY OF 25 M/S IN A DERICTION MAKING AN ANGLE OF 30 DEGREES UPWARD WITH THE HORIZONTAL

Answer 1

Explanation:

Assuming no air friction, etc., and recognizing that the derivative of vertical velocity is 0 when the projectile is at its maximum height, we get, by taking the derivative of y(t) = ½ gt^2 + (VoSinθ)t:

H = [(VoSinθ)2] / 2g

In this case, Vo = 20m/s and θ = 25°, so

H = (20 x 0.42262)^2 / (2 x 9.81) = 71.443/(2x9.81) = 3.641m

So, the expected height is 3.64m.