Question

An object is thrown up into the air with an initial velocity of 19.6 m/s, and caught again when it returns. Assuming it's acceleration is 9.81 m/s^2, how long does it take the object to reach its highest point?

Answer 1

The object's velocity at time t is given by the function

v(t) = 19.6 m/s - (9.81 m/s²) t

and at its highest point the object has zero velocity. Solve for t in this instance:

19.6 m/s - (9.81 m/s²) t = 0

t = (19.6 m/s) / (9.81 m/s²)

t ≈ 2.00 s