Question

velocity at the 3/4 of the maximim height where a=-g​

Answer 1

Explanation:

Maximum height: when the projectile is at the max ht,its vertical component of. velocity v=0. velocity u= usin. Distance (s) = H = max ht

1. accn(a)=-g
2. using v-u = 2as
3. 0-usinπ. =-2gH
4. H= usinπ/2g