1. Aug 8, 2010

### mheghan23

1.Figure shows an airplane moving horizontally with a constant velocity of 115 m/s at an altitude of 1050 m. The direction to the right and upward have been chosen at the positive directions. The plane releases a "care package". That falls to the ground along a curved trajectory. Ignoring air resistance determine the time required for the package to hit the ground.

2. For the same situation, find the speed of the package and the direction (angle) of the velocity vector. Just before the packages hits the ground.

horizontal motion
formula: X2= X1= Vx1t

Vertical motion
Y2=Y1+Vy1t - 0.5gt2

I Can't solve this because I don't how to solve this. My prof didn't gave us any clue for us to solve this problem..

3. The attempt at a solution

2. Aug 8, 2010

### arkofnoah

For 1, you just need to use the s = ut = 1/2at^2 formula to get the time taken to descend 1050. acceleration due to gravity is 9.81ms^-2, the initial vertical velocity is zero. plug in the values into the formula.

For 2, the horizontal velocity is the same. use the v = u + at formula to get the final vertical velocity and with a bit of geometry and vector sum you will get the direction and the resultant velocity.