Aprojectile is fired at an angle above the level ground. where does it have maximum kinetic energy? is it right when it leaves the launcher and right before it hits the ground?

You can approach this problem in an analytical way. The formula for kinetic energy is 1/2*mv². Throughout the projectile motion, mass is constant. Therefore, it wholly depends on the velocity. So, the maximum kinetic energy is when the object travels the fastest. This is right when it leaves the launcher, and the moment just before it hits the ground.

Distance covered = speed x timethe distance traveled going against the stream = distance traveled with the stream.therefore it is the speed and the time that are the variables.as he travels the same distance both ways, speed x time is the same for each way.going upstream his speed is 4 kph; going downstream his speed is 6 kph.therefore we have two simultaneous equations: time upstream x 4 = time downstream x 6time upstream + time downstream = 60 minutestherefore we have the equation: time upstream = 6/10 hour, time downstream therefore = 4/10 hour.distance traveled at 4km/hr in 6/10 hour = 24/10 = 2.4km.check: distance traveled at 6km/hr in 4/10 hour = 24/10 = 2.4km.