Question

A projectile is fired at time t = 0.0 s, from point 0 at the edge of a cliff, with initial velocity components of V ax= 50m/s and V ay= 200m/s. The projectile rises, then falls into the sea at point P. The time of flight of the projectile is 50.0s What is the y-coordinate of the projectile when its x-coordinate is 1000 m?

Answer 1

The projectile has constant horizontal speed all the time and we can use this formula

x=Vax*t, where t is time it (projectile) takes to reach x=1000m

t=x/Vax=1000/50=20s

As we know we can split projectile trajectory in two parts and the time for each part is equal. We can conclude that time t=20s belong to first part where projectile is going up.

Formula for this type of movement is  y=Vay-(gt∧2)/2

If we take acceleration of gravity g=10m/s∧2

we get     y=200*20-(10*20∧2)/2=4000-2000=2000m

But if I calculate time it takes the highest point I get  

t=Vay/g=200/10=20s

The highest point for that time is

y= H=Vay∧2/2g=200∧2/2*10=40000/20=2000m

This happened because the projecile is fired from the edge of a cliff which is above the sea line.

Good luck!!!