tin can problem - range of stream of water and max height above exit hole?

Question

A tin can is filled with water to a depth of 32 cm. A hole 11 cm above the bottom of the can produces a stream of water that is directed at an angle of 29° above the horizontal.
a)Find the range of this stream of water
b)Find the maximum height above the exit hole of this stream of water.
answer in meters please

Answer 1

To find the velocity of water exiting the hole use the fact that F=d(mv)/dt using P= pressure and A =area and density = 1
the volume of water that will exit per unit time is Av =d(m )/dt
and the force is F= PA =d(mv)/dt =v d(m )/dt =A v^2

Answer 2

fifty six*sin 50 = v(y) (speed y-course) v(y)^2=2*a*h h=ninety 3.seventy 9 upload 2.5 to get max correct = ninety six.29 m ninety 3.seventy 9+2.5 = .5a*t^2 t=4.40 3 fifty six*cos50 = v(x) v(x)*t = selection selection = 159.40 9 m