A man is in a canoe on a horizontal river going east at 500

A man is in a canoe on a horizontal river going east at 5.00 m/s, which turns into a waterfall with vertical (y-direction) drop/acceleration of -9.81 . After 2.00 seconds of falling from the river, he shoots a bullet in the eastward direction, which has a relative velocity with respect to the water of v = 60.0 m/s at an angle of 70 degrees above horizontal. Be careful, the waterfall has both x and y components of its velocity.

A) What is the magnitude of the bullet\'s initial velocity with respect to the ground?

B) How much time is required for the bullet to reach the highest point in its\' trajectory?

C) Suppose it takes 40 seconds to hit the ground. At what height was the gun shot?

D) How far has the bullet traveled horizontally from its inital position?

If you\'ve already answered this please do not answer. i want a second opinion.

Solution

A. velocity of the man in East( vx) = 5.0 m/s

    down ward accleration = 9.81 m/s

downward velocity after falling for 2.0 sec (vy) = 9.81*2 = 19.62 m/s

velocity of the bullet wrt water = 60 m/s

angle of firing = 70 deg above horizontal

horizontal component ux = 60 Cos(70) = 20.52 m/s

add to this water velocity vx

horizontal velocity wrt ground ugx = 20.52 +5.0 = 25.52

vertical component of the buller firing = 60Sin(70) = 56.38 m/s

water velocity is downard i.e. -y direction where as firing is in +y direction

vertical velocity wrt to ground ugy = 56.38-19.62 = 36.76

magnitude of the bullet initial velocity wrt groud ug^2 = 25.52^2 +36.76^2

   ug = 44.76 m/s

B. initial vertical velocity ugy = 36.76, the bullet will loose vel. due to gravity g=9.81

time to reach maximum height t = (ugy)/g =36.76/9.81 = 3.75 s

C. height after reaching the maximum h = u^2/2g = (36.76)^2/2*9.81 = 68.87 m

time of fall t = 40 -3.75   = 36.25 s

distance to fall = gt^2/2 = 9.81*36.25²/2 = 6445.48 m

The bullet has raised 68.87 m after firign and then descended 15696 m to hit the ground

height of firing from the ground = 6445.48 -68.87 = 6376.61 m

D. Total time of flight for the bullet = 40s

horizontal velocity = 25.52 m/s

horizontal dispalcement x = 25.52*40 = 1020.8 m