From an elevated point P, a stone is projected vertically upwards. When the stone reaches a distance h below P, its velocity is double of its velocity at a height h above P. The greatest height attained by the stone from the point of projection P is

Question

From an elevated point P, a stone is projected vertically upwards. When the stone reaches a distance h below P, its velocity is double of its velocity at a height h above P. The greatest height attained by the stone from the point of projection P is (A) (3/5)h (B) (5/3)h (C) (7/5)h (D) (5/7)h

Tardigrade Admin · Accepted Answer

From equations of motions, we know v2=u2-2 g h dots(i) Here, v=2 v, so (2 v)2=u2+2 g h dots(ii) Now, on adding Eqs. (i) and (ii), we get 5 v2=2 u2 u2=(5/2) v2 dots(iii) On subtracting Eq. (i) in Eq. (ii), we get 3 v2=4 g h v2=(4/3) g h dots(iv) And we know that the H=(u2/2 g) So, here H=((5/2) v2/2 g) =((5/2)((4/3) g h)/2 g) H =(5/3) h

Q. From an elevated point $P$ , a stone is projected vertically upwards. When the stone reaches a distance $h$ below $P$ , its velocity is double of its velocity at a height $h$ above $P$ . The greatest height attained by the stone from the point of projection $P$ is

$\frac{3}{5} h$

$\frac{5}{3} h$

$\frac{7}{5} h$

$\frac{5}{7} h$

$\frac{2}{3} h$

Q. From an elevated point P, a stone is projected vertically upwards. When the stone reaches a distance h below P, its velocity is double of its velocity at a height h above P. The greatest height attained by the stone from the point of projection P is

53​h

35​h

57​h

75​h