Question

a body is projected vertically up with certain velocity. At a point in its path

Answer 1

Kinetic energy = $\frac{1}{2} \times m \times v^2$

At point P,

Potential energy = mgH

Kinetic energy = $\frac{1}{2} \times m \times v_p^2$

Given,

$\frac{mgH}{\frac{1}{2}\times m \times v_p^2 } = \frac{9}{16}$

Kinetic energy at point P = $\frac{1}{2} \times m \times v^2$ - mgH

$\frac{1}{2} mv_p^2 + \frac{9}{32} mv_p^2 = \frac{1}{2}mv^2$

$\frac{25}{16}v_p^2 = v^2$

$\frac{v}{v_p}=\frac{5}{4}$

Hence, Ratio is 5 : 4

Correct option is B

Answer 2

Answer:

answer will be 5: 4

Explanation:

llllllll