Question

A ball 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

Answer 1

Answer:

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:

$v^2=u^2+2a(\Delta y)$

For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, $v=0\ m/s$

Also, $\Delta y=H=12\ m$

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

$0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s$

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

So, final velocity (v) = $\frac{u}{2}=\frac{15.34}{2}=7.67\ m/s$

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:

$v^2=u^2+2ah\\7.67^2=15.34^2+2\times -9.8\times h\\58.83=235.2-19.6h\\\\19.6h=235.2-58.83\\\\19.6h=176.37\\\\h=\frac{176.37}{19.6}\approx9.0\ m$

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