A particle is thrown vertically upwards. If its velocity at half of the maximum height is $10 ext{ m/s}$, th

Question

A particle is thrown vertically upwards. If its velocity at half of the maximum height is $10 	ext{ m/s}$, then maximum height attained by it is (Take $g = 10 	ext{ m/s}^2$):

Accepted Answer

1. **Define Maximum Height ($H$):** At maximum height, the final velocity $v = 0$. Let the initial velocity be $u$. Using the kinematic equation $v^2 = u^2 + 2as$ with $a = -g$:
   $$ 0^2 = u^2 - 2gH \implies u^2 = 2gH \quad \dots(1) $$
2. **Analyze Motion at Half Height ($H/2$):** The problem states that at height $s = H/2$, the velocity is $v = 10 	ext{ m/s}$. Apply the kinematic equation for this point:
   $$ v^2 = u^2 - 2g\left(\frac{H}{2}ight) $$
   $$ (10)^2 = u^2 - gH $$
3. **Substitute and Solve:** Substitute $u^2 = 2gH$ from equation (1) into the second equation:
   $$ 100 = 2gH - gH $$
   $$ 100 = gH $$
4. **Calculate $H$:** Given $g = 10 	ext{ m/s}^2$:
   $$ 100 = 10 	imes H $$
   $$ H = 10 	ext{ m} $$ .

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