Aball is being thrown straight upward and rises to a maximum height of 12.0m above its launch point. at what height above it’s launch point has the speed of the ball decreased to one-half of its initial value

9.0 m

Explanation:

Let the initial velocity be 'u'.

Given:

Final velocity is half of initial velocity.

Maximum height reached by ball (H) = 12.0 m

Acceleration of the ball is due to gravity (g) = -9.8 m/s²(Downward)

Now, first, we will find initial velocity of the ball using equation of motion given as:

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So,

Also,

Now, plug in all the values and solve for 'u'.

Now, consider the motion of the ball till the velocity reaches half of initial velocity.

So, final velocity (v) =

Now, again using the same equation and finding the new height now. Let the new height be 'h'.

So, equation of motion is given as:

Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.

the duration of a photographic flash is related to an rc time constant, which is 0.100 μs for a certain camera.