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