Answer:

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.


Initial speed of the projectile is v and final speed is 0.5 v.


g is the acceleration due to gravity
Let h is the height above the ground. Using the second equation of motion as :



So, the height of the projectile above the ground is
. Hence, this is the required solution.
Answer:
Distance =60m, Time = 6s, Speed = ?
Speed = distance/time
= 60/6
=10m/s
Explanation:
Hope that this is helpful.
Have a nice day.
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