# 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

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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 a*nd *-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__

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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)

⇒ t = [ 11.2 + 14.28]/(9.8)

⇒ t = 25.48/9.8 a

⇒ t = 2.6 second or -0.31 second

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