Free Fall physics

Question

A ball is thrown downward with initial velocity of 8.0m/s. What is it’s velocity 2.5s later? How far from its initial position it will be at that instant of time?

Answer 1

At constant acceleration,

 

v(t) = v₀ + at

 

Taking the downward direction as positive,

 

v₀ = 8.0 m/s

a = g = 9.8 m/s²

t = 2.5 s

 

v(t) = v₀ + gt = 8.0 + 9.8t  [m/s]

 

v(2.5) = 8.0 + 9.8(2.5) = 32.5 m/s in downward direction

 

Vertical fall distance is

 

Δy = gt²/2 + v₀t

 

You know g, v₀, and t.  Just plug in the numbers and get Δy.

Answer 2

v²-v0²=-wgy

v=v0-gt

y=-1/2gt²+v₀t

assuming g=10