Question

pls do this problem step by step

Answer 1

Answer:

50,25

Explanation:

1. Since it is a ground to ground projectile therefore when it is just about to hit the ground its velocity will be maximum and equal to initial velocity of 50m/s.
2. Since it is a ground to ground projectile,this means that the only acceleration acting on the projectile is g in downward direction which decreases the velocity of the projectile til it reaches the maximum height.At the maximum height the projectile has no y-component of its velocity that is , Vy =0 and since no air resist is observed along x therefore it has only ucos(60) as constant remaining velocity and hence
3. U=50 and Ucos(60) = 25 m/s
4. after crossing the maximum height point the g downward acceleration accelerates the projectile and increases the Vy component of its velocity while the Vx component remains constant so magnitude wise the velocity again increases.
5. so we arrive at the ans = 50ms^-1 => max velocity
6. and 25ms^-1 => min velocity
7. i hope you found the solution helpful please mark as brainliest if you did :)