How to find the distance and time of projectile motion?

You have two equations, one for the horizontal motion (x) and one for the vertical motion (y).  Here are the equations for the distance in each coordinate:

x:  x = v*cos?*t
y:  y = v*cos?*t - (1/2)*g*t^2 = 0

And you have the following givens:
v = 41.66 m/s
? = 45°
g = 9.81 m/s^2

You were also given the mass (m = 175 g), but this is not needed.

So... first start with the second equation, which was set equal to zero because it lands at the same height it was shot from, so total change in height is zero.  You only have one unknown, so you can solve for time.  Then, with that, you can plug the time into the first equation to solve for distance.

Finally, you will want how high it is going to go.  Well, the vertical motion is symmetric (i.e., it takes the same amount of time to go up as it does to come down).  So, after you solve for the total time of its flight, divide that in half.  This will be the time at its maximum height, so plug this into the second equation I gave you to solve for maximum height.

vx=41.66*1,4142/=29.46m/s
vy=29.46m/s
vertical motion
v^2-vy^2=-210h
h=vy^2/20=29.46^2/20=14.73m
vy=uy-10t.0=29.46-10t.t=2.95s
time of flight=2x2.95=5.90s
 horizontal distance=29.46x5.90=55.81m
horizontal motion