Velocity and acceleration of a ball thrown straight up

[spdrxc]

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?

 

[Wes Hughes]

(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

 

[H-bar None]

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.