Final answer:
Using the kinematic equations for uniformly accelerated motion and assuming the only acceleration is due to gravity, the time the ball remains in the air is found to be approximately 3.56 seconds, and its velocity just before striking the ground is approximately 34.9 m/s.
Step-by-step explanation:
To determine the time the ball remains in air and its speed just before striking the ground, we can use the kinematic equations for uniformly accelerated motion. Here, the acceleration 'a' is due to gravity (9.8 m/s2 since acceleration of the ball is not mentioned, we assume it's in free fall). Since the initial velocity 'vo' is zero (the ball is dropped), the two relevant equations are:
1. d = vot + ½ at2 (Distance travelled)
2. v = vo + at (Final velocity)
Insert the given values (d = 62.0 m, vo = 0, a = 9.8 m/s2) into the first equation to find time 't':
- 62.0 m = 0 + (½)(9.8 m/s2)t2
- t2 = (62.0 m)/(4.9 m/s2)
- t2 = 12.65
- t = √12.65
- t ≈ 3.56 s
Now, find the final velocity using the second equation:
- v = 0 + (9.8 m/s2)(3.56 s)
- v = 34.9 m/s
The time the ball remained in the air is approximately 3.56 seconds, and its velocity just before striking the ground is approximately 34.9 m/s.