A boat has a cannon on board that faces aft. The boat is capable of traveling at

Question

19 m/s on still water. The boat is traveling North on a river that has a current flowing south at a rate of 10 m/s. Following 20m directly behind the boat is a raft that is traveling at 13 m/s relative to the water. On the raft is a clown with a hoop. The cannon fires the cannonball at 38.5 with velocity 20 m/s. At what velocity must the clown throw the hoop in order for the cannonball to travel through it? Assume the clown throws the hoop straight up.

Answer 1

If the elevation angle of the cannon is 38.5 degrees
then the horizontal velocity of the cannon ball relative to earth is
u = 19 - 10 - 20cos 38.5 forever
Its location relative to earth from the time of firing is
x = u t

the initial vertical velocity of the cannon ball is
Vi = 20 sin 38.5
its vertical velocity is
v = 20 sin 38.5 - 9.81 t
its height is
h = Vi t - 4.9 t^2

now the hoop
the horizontal velocity of the hoop relative to the ground is 13-10 = 3 m/s
the x position of the hoop is
xh = -20 + 3 t
the vertical speed of the hoop is
vh = Vhi - 9.81 t
the height of the hoop is
Hh = Vhi t - 4.9 t^2

now we want
x = xh
and
h = Hh

Answer 2

u = -6.65 m/s

u t = -20 + 3 t
20 = (3 + 6.65)t
t = 2.07 secs

how high?
h = 12.45(2.07) - 4.9(2.07)^2
= 25.8 - 21 = 4.8 meters
so
4.8 = Vih (2.07)-4.9(2.07)^2
Vih = 12.5 meters/ second
check my arithmetic !