The equation that relates distance, velocities, acceleration, and time is, d = V₀t + 0.5gt² where d is distance, V₀ is the initial velocity, t is time, and g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock, (3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²) The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s (3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²) The value of t from the equation is 5.61 s
The difference between the tim, difference = 24.73 s - 5.61 s difference = 19.12 s
you can figure out the time that the dropped ball to hit the ground quite easily because u = 0 m/s, therefore
s = (1/2)at²
t² =2s/a
t = sqrt(2s/a)
assuming the acceleration from gravity is 9.81 m/s²
t = sqrt(2*300m/9.81 m/s²)
t = 7.8 s
You can use the quadratic equation to find t for the thrown rock and then subtract them. Or you can calculate the final velocity of the thrown rock with v^2 = u^2 + 2as and then use v = u + at to find the time for the thrown rock.
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