Help with physics problem please?

Question

A tennis player serves a ball horizontally. What minimum speed is required for the ball to clear the 0.90m high net about 15.0 from the server if the ball was launched from a height 2.5m above the ground? where will the ball land if it just clears the net? How long in the air assuming air resistance is negligible?

What I did so far was subtract the height of the net from the height of the person. So i get

x0 = y0 = 0

x = 15

y = -1.6

ax= 0

ay= -9.8

vy0= 0

vx0= ??

t= ??

should i just treat it as if its a ball off a cliff now? and what about the second part...thank you

Answer 1

**first of all i prefer assuming g=10 not 9.8 to ease calculation.**

(The Min speed):

horizontally:

speed=u=?               time=?          dist=15

vertically:

initial velocity(u)=0               acceleration(a)=10  

**I used 10 not -10 and the displacement will be 1.6**

displacement(S)= 2.5-0.9=1.6          time(t)=??

S=ut+0.5at^2

  =0+0.5*10*t^2         so time=0.57 sec

back to horizontal:

now Speed=?  time=0.57 from vertical calculations

distance=15 so speed=26.3 m/s

(Place of landing):

**when it lands so the vertical distance is 2.5**

vertical:

u=0      a=10               s=2.5            t=?

S=ut+0.5at^2

2.5=0.5*10*t^2        so t=0.707 sec

Horizontal:

speed=u=26.3                      time=0.707

distance=18.6m    "after the net by 3.6m"

the last part is not clear how long in the air what.....u mean stay in the air if so then the answer is the total flight =0.707 s