Also if its possible part B wants to know A ball thrown vertically upward is caught bythe thrower after 2 s.Find the initial velocity of the ball. Theacceleration of gravity is 9.8 m/s2Answer in units of m/sFind the maximum height it reaches.Answer in units of m
A) Speed of falling brush = at =9.8 x 3.39 = 33.22m/sec B) Vertical distance Y = Vo x t - 0.5 x at^2 0 = Vo x 2 - 0.5 x 9.8 x 4 Initial Velocity = Vo = 9.8m/sec^2 Vertical height = Vo^2/2g = 9.8^2/2 x 9.8 = 4.9m
Use, Vf = Vo + g * t With Vo = 0 , Vf = g * t = 9.8 m/s^2 * 3.39 s = 33.222 m/s. part B, use the same equation, but take into consideration to watch for direction and adopt a convention. As before we have considered +ve for downward direction. Now Vf = 0 0 = Vo + g * t. Vo = - 9.8 m/s^2 * 2 s = -19.6 m/s . Negative means upward direction according to the adopted convention.
