Final answer:
The speed of the airplane releasing a ball that lands 235m from the release point can be calculated using the equations of projectile motion, accounting for the time it takes for the ball to fall to the ground and the ball's horizontal displacement.
Step-by-step explanation:
To calculate the speed v of the airplane from which a ball is released and lands 235m away from the release point, we can use the kinematic equations for projectile motion. We know the ball falls 235 meters in the horizontal direction, and we have the height from which it is dropped, 235m, as well as the acceleration due to gravity, 9.81m/s2.
We'll use the following kinematic equation for the vertical (y) motion, where u_y is the initial vertical speed (which is 0), s_y is the vertical distance, a_y is the acceleration due to gravity, and t is the time:
- s_y = u_y * t + 0.5 * a_y * t2
Solving for t (time of fall) gives:
Substitute the given values:
- t = √(2 * 235m / 9.81m/s2)
Once we have the time, we can use the horizontal (x) motion to find the initial velocity v of the airplane, as the horizontal speed of the ball equals the speed of the plane:
Solve for v:
We then plug in the values for s_x (235m) and t (the time we found previously), and compute for v, the speed of the plane.