Answer:
The speed is 20.8 m/s
Step-by-step explanation:
If a projectile ascends vertically, and after 3 seconds it reaches its maximum height, calculate the velocity that it carries to the middle of its downward trajectory
Let the maximum height is h and initial velocity is u.
From first equation of motion
v = u + at
0 = u - g x 3
u = 3 g.....(1)
Use third equation of motion
[tex]v^2 = u^2 - 2 gh \\\\0 = 9 g^2 - 2 gh \\\\h = 4.5 g[/tex]
Let the speed at half the height is v'.
[tex]v^2 = u^2 + 2 gh \\\\v'^2 = 0 + 2 g\times 2.25 g\\\\v'^2 = 4.5\times 9.8\times9.8\\\\v' = 20.8 m/s[/tex]
