Answer:
[tex]h=\dfrac{3v^2}{8g}[/tex]
Explanation:
It is given that,
Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.
[tex]v=u+at[/tex]
[tex]v=u-gt[/tex]
Initial speed of the projectile is v and final speed is 0.5 v.
[tex]0.5v=v-gt[/tex]
[tex]t=\dfrac{v}{2g}[/tex]
g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :
[tex]h=vt-\dfrac{1}{2}gt^2[/tex]
[tex]h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2[/tex]
[tex]h=\dfrac{3v^2}{8g}[/tex]
So, the height of the projectile above the ground is [tex]\dfrac{3v^2}{8g}[/tex]. Hence, this is the required solution.
