From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is double of what it was at a height h above A. Show that the greatest height attained by the stone is $\frac{5}{3} h .$

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Solution

Let u be the velocity with which the stone is projected vertically upwards. Given that, $v_{- h} = 2 V_{h}$
or ${(v_{- h})}_{- h}^{2} = 4 v_{h}^{2}$
or $u^{2} - 2 g - (- h) = 4 (u^{2} - 2 g h)$
$∴ u^{2} = \frac{10 g h}{3}$
Now, $h_{m a x} = \frac{u^{2}}{2 g} = \frac{5 h}{3}$

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From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is double of what it was at a height h above A. Show that the greatest height attained by the stone is 53h.

Let u be the velocity with which the stone is projected vertically upwards. Given that, v−h=2Vhor (v−h)2=4v2hor u2−2g−(−h)=4(u2−2gh)∴u2=10gh3Now, hmax=u22g=5h3

From an elevated point A, a stone is projected vertically upwards. When the stone reaches a distance h below A, its velocity is double of what it was at a height h above A. Show that the greatest height attained by the stone is $\frac{5}{3} h .$

Let u be the velocity with which the stone is projected vertically upwards. Given that, $v_{- h} = 2 V_{h}$
or ${(v_{- h})}_{- h}^{2} = 4 v_{h}^{2}$
or $u^{2} - 2 g - (- h) = 4 (u^{2} - 2 g h)$
$∴ u^{2} = \frac{10 g h}{3}$
Now, $h_{m a x} = \frac{u^{2}}{2 g} = \frac{5 h}{3}$