Respuesta :
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:
[tex]v^2=u^2+2a(\Delta y)[/tex]
For maximum height, final velocity is 0 as the ball stops at the maximum height temporarily. So, [tex]v=0\ m/s[/tex]
Also, [tex]\Delta y=H=12\ m[/tex]
Now, plug in all the values and solve for 'u'.
[tex]0^2=u^2+2(-9.8)(12)\\\\u^2=235.2\\\\u=\sqrt {235.2} =15.34\ m/s[/tex]
Now, consider the motion of the ball till the velocity reaches half of initial velocity.
So, final velocity (v) = [tex]\frac{u}{2}=\frac{15.34}{2}=7.67\ m/s[/tex]
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:
[tex]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[/tex]
Therefore, the height reached by the ball when velocity is decreased to one-half of the initial velocity is 9.0 m.
