Calculating Ball Velocity: Understanding the Movement of a Thrown Ball

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SUMMARY

The discussion focuses on calculating the velocity and acceleration of a ball thrown vertically. At its highest point, the ball's velocity is 0 ft/sec, confirming that it has reached its peak. One second before reaching this point, the velocity is 32 ft/sec upward, while one second after, it is 32 ft/sec downward. The change in velocity during the 1-second intervals before and after the peak is consistently 32 ft/sec, with a constant acceleration of 32 ft/sec² downward throughout the ball's flight.

PREREQUISITES
  • Understanding of basic physics concepts, specifically kinematics
  • Knowledge of gravitational acceleration (32 ft/sec²)
  • Familiarity with velocity and acceleration definitions
  • Ability to perform basic calculations involving time intervals and changes in velocity
NEXT STEPS
  • Study kinematic equations for uniformly accelerated motion
  • Learn about projectile motion and its applications
  • Explore the effects of air resistance on ball trajectory
  • Investigate the relationship between launch angle and maximum height
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Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of motion related to thrown objects.

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A ball is thrown with enough speed straight up so taht it is in the air for several seconds. (a) What is the velocity of the ball when it gets to its highest point? (b) What is the velocity 1s before it reaches its highest point? (c) What is the change in velocity during this 1s interval? (d) What is its velocity 1s after it reaches its highest point? (e) What is the change in velocity during this 1s interval? (f) What is the change in velocity during the 2s interval? (g) What is the acceleration of the ball during any of these time intervals and at the moment the ball has zero velocity?

I am completely and utterly confused. Can someone please help me walk through this?
 
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(a) When the ball gets to the highest point -- its speed or velocity is 0 (zero.) This must be the case or else the ball would go higher and it would not be at it's highest point yet.

(b) Since the accelleration is 32 ft per sec. per sec downward it's velosity would be 32 ft per sec upward, so that in 1 sec it's velosity would be 0.

(c) 32 ft per sec. This is standard gravity.

(d) 32 ft / sec down.

(e) 32 ft /sec down.

(f) 64 ft / sec

(g) 32 ft per sec per sec downward Always!

Hope this helps Wes Hughes
 
I thought I'd chime in a little.

The key to all this is to remember the ball's acceleration is constant. It doesn't matter where the ball is at during its flight. It doesn't matter how hard the ball is launched.
 

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