final speed and time of a dropped ball?

Question

A 4.12 kg ball is dropped from the roof of a 170.9 m tall building. While the ball is falling to Earth, a horizontal wind exerts a constant force of 13.3 N on the ball.

The acceleration of gravity is 9.81 m/s^2
A) How long does it take to hit the ground (seconds)?
B) What is its speed when it hits the ground? (m/s)

Answer 1

The vertical motion, assuming zero initial velocity is h - .5*g*t^2. Regardless of the horizontal acceleration, the time it takes to reach the ground is given by that formula. Thus you can calculate t for the answer to A. The speed at which it hits the ground will have two components, vertical and horizontal. The verical speed vy = g*t, where t is what you just calculated. The horizontal motion of the ball is given by f = m*a, or a = f/m. The horizontal speed is vx = a*t; t is tha same value as before, you have f and m, so you can calculate vh. The total velocity is the vector sum of the two (vx and vy) which are at right angles to each other, therefore the ball's total velocity is √(vx^2 + vy^2).

I thought it better to give you the way to find the answer rather than the numbers.

Answer 2

5.903 sec
60.96 m/sec