solving projectile problems

Question

 A ball is launch with the horizontal velocity of 10.0 m/s from a 20.0 m tall cliff.  how long will it be in the air? And how far will it land from the base of the cliff? 

Answer 1

Ignoring air resistance, and with no initial vertical velocity, the ball will fall from height h to the ground in time t which we can get from

 

h = (1/2)gt²

 

t = √(2h/g) = √(2*20.0/9.81) s ≅ 2.02 s

 

That is when it hits the ground.  So the horizontal distance that it travels is

 

x = v_0xt = (10.0 m/s)(2.02 s) = 20.2 m