Question

Plz solve 15,16and17 questions

Answer 1

▶ Solution :

✴ Given :

▪ A ball is thrown up vertically with speed u. At the same instant another ball B is released from rest at a height h vertically above the point of projection.

▪ Taking the moment of projection as t=0 and acceleration due to gravity is g.

✴ To Find :

1) At the moment of collision, the position of A is at a distance...

2) At the moment of collision, the position of B is at a distance...

3) The time instant at which they collide is

✴ Assumption :

✏ Let both meet at a distance x from height (h) and (h-x) from the ground at time t.

✴ Calculation :

⏭ Distance covered by ball B in time t

$\implies\sf\:x=\dfrac{1}{2}gt^2\\ \\ \therefore\bf\:t=\sqrt{\dfrac{2x}{g}}$

⏭ Distance covered by ball A in time t

$\implies\sf\:h-x=ut-\dfrac{1}{2}gt^2\\ \\ \dag\sf\:putting\:value\:of\:t\\ \\ \implies\sf\:h-x=u\times\sqrt{\dfrac{2x}{g}}-\dfrac{1}{2}g\times\dfrac{2x}{g}\\ \\ \implies\sf\:h-x=u\sqrt{\dfrac{2x}{g}}-x\\ \\ \implies\sf\:h=u\sqrt{\dfrac{2x}{g}}\\ \\ \implies\boxed{\bf{\pink{x=\dfrac{gh^2}{2u^2}}}}\\ \\ \implies\boxed{\bf{\purple{h-x=h-\dfrac{gh^2}{2u^2}}}}\\ \\ \dag\sf\:putting\:value\:of\:x\:in\:eq\:of\:t\\ \\ \implies\sf\:t=\sqrt{\dfrac{2x}{g}}\\ \\ \implies\sf\:t=\sqrt{\dfrac{2}{g}\times \dfrac{gh^2}{2u^2}}\\ \\ \implies\boxed{\bf{\green{t=\dfrac{h}{u}}}}$

15) → 1

16) → 4

17) → 4