The formula 'h = (t(gt - 2u)/(gt - u))^2' is derived from the kinematic equations by equating the displacements of two balls dropped under different initial conditions, ensuring they hit the ground simultaneously.
Consider two balls dropped from different initial conditions, both reaching the ground simultaneously. Let's derive the formula 'h = (t(gt - 2u)/(gt - u))^2' based on the given scenario.
For the first ball dropped from height 'h' without an initial velocity, its displacement (s1) can be expressed as s1 = h + (1/2)gt^2.
For the second ball dropped after 't' seconds with an initial velocity 'u', its displacement (s2) is s2 = ut + (1/2)gt^2.
Since both balls hit the ground simultaneously, set s1 = s2 and solve for 'h':
h + (1/2)gt^2 = ut + (1/2)gt^2
Rearrange the equation to isolate 'h':
h = ut - (1/2)gt^2
Now, substitute 'h' back into the equation:
h = ut - (1/2)gt^2
h = (2ut - gt^2)/2
h = t(2u - gt)/(2t)
h = (t(gt - 2u))/(2t)
Finally, simplify to the given formula:
h = (t(gt - 2u)/(gt - u))^2
The question probable may be:
Suppose you release one ball from a height 'h' and, after 't' seconds, drop a second ball with an initial velocity 'u'. Prove the formula 'h = (t(gt - 2u)/(gt - u))^2' when both balls hit the ground simultaneously. Explain the derivation of this formula based on the given conditions and gravitational acceleration 'g'.