An object is thrown vertically upward with an initial velocity of 40m/s. Two seconds later another object is thrown upward with the same velocity. At what height they will meet?

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Solution

Dear Student,
Let both the objects meet at h after a time t.
Height traversed by 1st object at time t:
$h = u t - \frac{1}{2} g t^{2}$ .......(1)
Height traversed by 2nd object at time (t-2)
$h = u (t - 2) - \frac{1}{2} g {(t - 2)}^{2} = u t - 2 u - \frac{1}{2} g (t^{2} - 4 t + 4) = u t - 2 u - \frac{1}{2} g t^{2} - 2 g t + 2 g$ .......(2)
Comparing eq. (2) with eq. (1), we get
$u t - \frac{1}{2} g t^{2} = u t - 2 u - \frac{1}{2} g t^{2} + 2 g t - 2 g - 2 u + 2 g t - 2 g = 0 t = \frac{2 u + 2 g}{2 g} = \frac{2 \times 40 + 2 \times 10}{2 \times 10} (t a k i n g g = 10 m s^{- 2}) t = 5 s$
Height traversed in 5 s
$h = u t - \frac{1}{2} g t^{2} = 5 \times 40 - \frac{1}{2} \times 10 \times 5^{2} = 200 - 125 = 75 m$
Regards

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