Question

Q] Explain every step briefly.
1) The maximum height reached by a projectile is 4 m. The horizontal range is 12 m. Velocity of projection in $ms^{-1}$ is   (g is acceleration due to gravity)
{ answer should come 5√g/2 }

2)A stone is thrown at an angle Фto the horizontal reaches a maximum height H. Then the time of flight of stone will be:          { answer should come 2√H/g }

Answer 1

we remember these formulas:

$Max\ height = H = (u\ Sin \theta)^2 / 2 g , \\ \\ Range = R = (u Cos \theta)(2 u\ Sin \theta)/g = u^2 Sin\ 2\theta/g \\ \\ H/R = Tan\ \theta / 4$

$H = u^2 Sin^2\ \theta /2g = 4 m, \ \ \ \ \ R = u^2 Sin\ 2\theta /g = 12m \\ \\ H/R= tan\ \theta/4 = 4/12 \\ \\ tan\ \theta = 4/3,\ \ \ sec\ \theta = 5/3,\ \ \ cos\ \theta= 3/5,\ \ sin\ \theta=4/5 \\ \\ u^2 = 2 g H /sin^2\ \theta = 2 g 4 / (4/5)^2 = 25g/2 \\ \\ u = 5\sqrt{g/2}$
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$v = u + at, \ \ 0 = u Sin\ \theta - g \ t \\ \\ t = u sin\ \theta/g \\ Distance\ traveled\ H\ = u t + \frac{1}{2} g t^2 = u * u sin\ \theta/g - 1/2 g (u sin\ \theta/g)^2\\ = u^2\ sin^2\ \theta/2g, \ it\ is\ same\ as\ 1/2 gt^2 \\ \\ So\ t = \sqrt{2H/g} \\ \\ The\ total\ time \of\ flight\ for\ ascending\ and\ descending = 2 * \sqrt{2H/g}$

Time of flight = time to reach the top + time to descend. Each is = root(2H/g) = u Sin Ф/g

Note: (there is 2 root 2.)