Question

Two balls of the same mass are thrown
upwards at an interval of 2 seconds, along the
same vertical line with the same initial
velocity of 39.2 ms-!. They will collide at a
height​

Answer 1

Answer:

$\huge\mathfrak\red{answer :-}$

Let the two balls collide t sec after the first ball is thrown, and let h be the height at which they collide.

For the first ball:

$h = 40t - \frac{1}{2} g {t}^{2}$

And for the second ball:

$h = 40(t - 2) - \frac{1}{2} g(t - {2)}^{2}$

$40t - \frac{1}{2} g {t}^{2} = 40(t - 2) - \frac{1}{2} g(t - {2)}^{2}$

$2gt = 80 + 20 = 100$

$t = 5sec$

$\therefore \: h = 40 \times 5 - \frac{1}{2} \times 10 \times {5}^{2}$

$= 75m$

${\huge{\boxed{\mathcal{{h = 75m}}}}}$

Explanation:

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- Tanvi.

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Answer 2

Explanation:

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