A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertically up from the ground with an initial velocity of 0=40.0 m/s . At what height from the ground will the two objects first meet?

33.23 m

Explanation:

At the point where both objects will meet, the vertical height will be equal.

From the equations of motion, the vertical height of the body falling at any time is given as

(y - y₀) = ut + ½gt²

y = vertical height at any time T

y₀ = initial height of the object = 81.5 m

u = initial velocity = 0 m/s (body was dropped)

g = -9.8 m/s²

(y - 81.5) = 0 - 4.9T²

y = 81.5 - 4.9T² (eqn 1)

For an object thrown up, the vertical height of the body at any time, t, is given as

(y - y₀) = ut + ½gt²

y = vertical height of the object at any time t

y₀ = initial height of the object = 0 m

u = initial velocity = 40 m/s

g = -9.8 m/s²

y = 40t - 4.9t² (eqn 2)

At the point where the two objects meet, we equate eqn 1 and eqn 2

y = y

81.5 - 4.9T² = 40t - 4.9t²

But T = (t + 2.2) (Since object 2 was dropped 2.2 s after object 1)

81.5 - 4.9(t + 2.2)² = 40t - 4.9t²

81.5 - 4.9(t² + 4.4t + 4.84) = 40t - 4.9t²

81.5 - 4.9t² - 21.56t - 23.716 = 40t - 4.9t²

81.5 - 21.56t - 23.716 - 40t = 0

57.784 = 61.56t

t = (57.784/61.56) = 0.93866 = 0.94 s

Therefore, the vertical height at t = 0.93866 s is

y = (40×0.93866) - 4.9(0.93866²) = 33.23 m

Hope this Helps!!!

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