Question

Imagine that you throw a ball vertically upwards from the rooftop of a building. The ball abandons your hand at a point that has the same height as the railway of the rooftop with an ascending speed of 15m/s, been left in free falling. Coming down the ball crosses the railway, at this point gravity is 9.81 m/s2.
Calculate the displacement and velocity of the ball 1 s and 4 s after been let go.
b)The velocity when the ball is 5 m above the railway.
c) Maximum height reached and the moment in which the ball reaches that height
The acceleration of the ball at its maximum height

Answer 1

a. D = Vo*t + 0.5g*t^2
D = 15*1 - 4.9*1^2 = 10.1 m.
V = Vo + g*t = 15 - 9.8*1 = 5.2 m/s

Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise time.

h max = -(Vo^2)/2g = -(15^2)/-19.6 = 11.48 m. Above launching point.

Tf = 4 - 1.53 = 2.47 s.

D = 4.9*Tf^2 = 4.9*2.47^2 = 29.9 m After
4 s.
V^2 = Vo^2 + 2g*d = 0 + 19.6*29.9 = 585.93
V = 24.2 m/s.

b. V^2 = Vo^2 + 2g*(11.48-5) = 0 + 127
V = 11.27 m/s.

c. h max = 11.48 m.(See previous calculation).

Tr = 1.53 s.(See previous calculation).