Question

If a ball is thrown upwards at a speed of 11.2 km/s from the balcony, 4m above the ground, how much time would it take to strike the ground at the base of

Answer 1

There is a mistake in your question. The question should be "a ball is thrown upwards at a speed of 11.2 m/s from the balcony", not  11.2km/s. If you throw something at 11.2 km/s from earth's surface, it will never return to ground.
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For solving the question, I am assuming upward direction as positive and ground as zero elevation.

initial velocity, u = +11.2 m/s  
initial position, s₀ = +4 m
g = -9.8 m/s²  (as g acts downward.)

when the ball strikes the ground,
final position, s₁ = 0 m
displacement, s = s₁ - s₀ = 0 - 4 = -4m
let the time taken = t

Using the equation of motion,
s = ut + 1/2 gt²
⇒ -4 = 11.2×t + 1/2 × (-9.8) × t²
⇒ -4 = 11.2t - 4.9t²
⇒ 4.9t² - 11.2t -4 = 0

Solving the above equation
t = [ 11.2 + √(11.2²+4×4×4.9)]/(2×4.9) and  [ 11.2 + √(11.2²+4×4×4.9)]/(2×4.9)
⇒ t =  [ 11.2 + 14.28]/(9.8) and  [ 11.2 - 14.28 ]/(9.8)
⇒ t = 25.48/9.8 and -3.08/9.8
⇒ t = 2.6 second or -0.31 second

since time can't be negative, ignore t=-0.31s

So time taken by the ball to strike the ground is 2.6s