A ball is thrown upward with a speed of 40 m/s. Approximately how much time does it take the ball to travel from the release location (A). Till its highest point (B)? Approximately how much total time is the ball in the air before it returns back to its original height (C)?

I'm going to assume that this gripping drama takes place on planet Earth, where the acceleration of gravity is 9.8 m/s².  The solutions would be completely different if the same scenario were to play out in other places.

A ball is thrown upward with a speed of 40 m/s.  Gravity decreases its upward speed (increases its downward speed) by 9.8 m/s every second.

So, the ball reaches its highest point after (40 m/s)/(9.8 m/s²) = 4.08 seconds. At that point, it runs out of upward gas, and begins falling.

Just like so many other aspects of life, the downward fall is an exact "mirror image" of the upward trip.  After another 4.08 seconds, the ball has returned to the height of the hand which flung it.  In total, the ball is in the air for 8.16 seconds up and down.

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

the answer is a. hope this .