Question

. A particle projected vertically upwards attains a maximum height H. If the body takes
one-third of time of ascent to reach a height h (h < H) then
(A) 4h = 3H (B) 4H = 3h C) 3h = H (D) 4h = H​

Answer 1

Answer:

Find the velocities at h

Since

$t = \frac{u}{g}$

one third means

$t = \frac{u}{3g} \\ v = u - gt = \frac{2u}{3}$

Now

$s = \frac{ {v}^{2} - {u}^{2} }{2g}$

So ratio will be

$\frac{H}{h} = \frac{0 - {u}^{2} }{( \frac{2u}{3} ) {}^{2} - {u}^{2} } = \frac{9}{5}$

Answer 2

Answer:

Step-by-step explanation:

Find the velocities at h

Since

t = \frac{u}{g}

one third means

t = \frac{u}{3g} \\ v = u - gt = \frac{2u}{3}

Now

s = \frac{ {v}^{2} - {u}^{2} }{2g}

So ratio will be

\frac{H}{h} = \frac{0 - {u}^{2} }{( \frac{2u}{3} ) {}^{2} - {u}^{2} } = \frac{9}{5}