Cannonball fired at 45°: finding maximum height, time of flight and range

[F.B]

I need help on projectile motion questions because i had to miss a whole lesson for some reason and my teacher said to try some of the questions.
Anyways I am stuck on this question.

A cannon is set at an angle of 45 degrees above the horizontal. A cannon ball leaves the muzzle with a speed of 220 m/s. Air resistance is negligible. Determine the cannonball's
a) maximum height
b) time of flight
c) horizontal range (to the same vertical level)

This is what i did. I am not going to do step by step on the part i got.

Before these i started i found VIx which is 155.6 m/s and Viy is the same
a) At max height Vfy=0( i don't get why but i know that's how you do it)
2ad=Vi^2 + Vf^2
d= (155.6)^2/-2(-9.8)
d= 1240m

b)i have no idea how to get this one i hope you can help me.

c)i started but i couldn't finish. This is what i did

d=Vit + 1/2a(t^2)
1240=155.6t - 4.9t^2
4.9t^2 - 155.6t + 1240 = 0

I know i need to do the quadratic formula after this but i get a negative with the radical so i don't know what to do i think i made a mistake somewhere.

Can you please help me?

 

[Integral]

To understand why the v_y=0 at the peak of motion you need to think about how gravity is effecting the motion. Gravity is always acting toward the earth, so as long as the projectile has a upward directed velocity (V_y>0 it will be slowed by gravity. Now consider what happens when it has been slowed by gravity to the point it has lost all of its upward motion. It will stop, so V_y=0, now it will begin to fall, so now V_y<0 . Can you figure out from this why V_y=0 is at the highest point?

 

[hotvette]

For b, you need to think about what the velocity components are when the flight is finished.

For c, if you have an equation for distance vs time, the answer to b will be useful.

hotvette