Answer :
The time to reach the highest point is 3 seconds.
To solve the given questions, we can use the kinematic equations of motion. Let's go through each question one by one:
The velocity of the ball 1 s after launch,
The initial velocity of the ball is -66 mph (negative sign indicating upward direction).
The acceleration due to gravity is approximately 22 mph per second (also in the upward direction).
Using the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can calculate:
v = -66 mph + (-22 mph/s) * 1 s = -66 mph - 22 mph = -88 mph.
The velocity of the ball 2.5 s after launch,
Using the same equation of motion, we can calculate:
v = -66 mph + (-22 mph/s) * 2.5 s = -66 mph - 55 mph = -121 mph.
What is the velocity of the ball 3 s after launch?
Using the same equation of motion:
v = -66 mph + (-22 mph/s) * 3 s = -66 mph - 66 mph = -132 mph.
the velocity of the ball 4.5 s after launch,
Using the same equation of motion:
v = -66 mph + (-22 mph/s) * 4.5 s = -66 mph - 99 mph = -165 mph.
the velocity of the ball 5.5 s after launch,
Using the same equation of motion:
v = -66 mph + (-22 mph/s) * 5.5 s = -66 mph - 121 mph = -187 mph.
the velocity of the ball 6 s after launch,
Using the same equation of motion:
v = -66 mph + (-22 mph/s) * 6 s = -66 mph - 132 mph = -198 mph.
How long does it take the ball to reach the highest point?
The ball reaches the highest point when its velocity becomes zero. Using the equation v = u + at, and setting v = 0 mph, we can solve for t:
0 = -66 mph + (-22 mph/s) * t_highest.
Solving for t_highest, we find:
t_highest = 66 mph / 22 mph/s = 3 seconds.
How long does it take the ball to return back down to the same height?
Since the time to reach the highest point is 3 seconds, it will take the same amount of time to return back down to the same height.
Therefore, 3 seconds.
Learn more about time from the given link
https://brainly.com/question/479532
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