A ball is thrown upward and reaches a height of 3 meters. What is the initial velocity of the ball and the time to reach the 3 meters?

Take g = 9.81 m/s²
Take all upward quantities as positive.

Initial velocity, u = ? m/s
Final velocity, v = 0 m/s
Acceleration, a = -9.81 m/s²
Time taken, t = ?
Displacement, s = 3

v² = u² + 2as
0 = u² + 2(-9.81)(3)
Initial velocity, u = √[2(9.81)(3)] = 7.7 m/s

(1/2)(v + u)t = 3
(1/2){√[2(9.81)(3)] + 0}t = 3
t = 6/√[2(9.81)(3)] = 0.78 s

v = u - gt ..............(1) vertical velocity

h = ut - 1/2gt^2 .....(2) vertical height

At max h,

u - gt = 0

t = u/g

3 = u(u/g) - 1/2g(u/g)^2

= u^2/g - u^2/2g


= u^2/2g

u = sqrt[3(2g)]

= 7.668115805

= 7 . 67 m/s

t = 7.67/9.8

= 0.782467964

= 0 , 78 s

Initial velocity is 7.668m/s
Time= 0.782s