Final answer:
The time the ball remains in the air after being dropped from a height of 72.0m is approximately 3.8 seconds, and the ball's speed just before striking the ground is approximately 37.3 m/s.
Step-by-step explanation:
The student's question is about calculating the time a ball remains in the air when dropped from a height and its speed just before striking the ground. We can use kinematic equations for an object in free fall to solve this.
Time of Flight
To find the time the ball remains in the air, we use the equation:
s = ut + (1/2)at^2
Here, s is the distance the ball falls (72.0 m), u is the initial velocity (0 m/s, since the ball is dropped), a is the acceleration due to gravity (9.81 m/s^2), and t is the time in seconds. Plugging in the values we find that t^2 is approximately 14.7 seconds squared, hence t is approximately 3.8 seconds.
Final Speed Before Impact
The final speed v just before striking the ground is given by:
v = u + at
Using the values for u, a, and t we previously found, we get that the ball's final speed just before impact is approximately 37.3 m/s.