A ball is thrown up with an initial speed of 80m/s. The speed of the object after 11 second is?

Answer 1

1) d^2y/dt^2 = -9.81

*integrate*

2) dy/dt = -9.81t + C

--> to determine C, we know that dy/dt=80 m/s for t=0

3) dy/dt = -9.81t + 80

Answer 2

From kinematics,

v(t) = v₀ - gt

v₀ = 80 m/s

g = 9.8 m/s²

v(11) = ?

Plug in the numbers and finish from here.