Question 1178129

Here acceleration (a) is g = 9.8m/s^2 (this will be negative since it is going in the opposite direction of the initial velocity),

initial velocity(u)  = 15m/s
final velocity (v) = 0m/s if the ball stops
v^2 = u^2 + 2a*h, where h is the max height of the ball,
 we can find h by solving: h = (v^2 - u^2)/2a,
 we plug in  h = (0^2 - 15^2)/2(-9.8) = 11.48 m.

b)calculate how long the ball is in the air before it comes back to his hand.

 time to reach its peak and back down to our hand.
 Using this, the time it takes to get to the top * 2 = total time of flight . Using d = (v + u)*t / 2, ) = 2(11.48)/(15) = 1.53 s to reach top,

we get a total of2*1.53 = 3.06 s to go up and back down to our hand.