First slide

Rectilinear Motion

Question

A body is thrown vertically upward. At half of the maximum height, the velocity of the body is 10m/s. The maximum height reached by the body is

Moderate

A

10 m
B

15 m
C

20 m
D

0 m

Solution

Let ‘u’ be the initial velocity
$∴$ At half of the maximum height, let ‘v’, be the final velocity.
$∴$ From equation $v^{2} - u^{2} = 2 a s$
$\Rightarrow v^{2} - u^{2} = - 2 g (\frac{h}{2}) ......... (i)$
$v_{1} = 10 m / s$
We know that maximum height reached by the body is
$\begin{array}{l} h = \frac{u^{2}}{2 g} \\ \Rightarrow u^{2} = 2 a h \\ ∴ From (i) 100 - (2 g h) = - g h \\ \Rightarrow g h = 100 \\ ∴ h = \frac{100}{g} = \frac{100}{10} = 10 m \end{array}$
II method: At highest point v=0
$v^{2} - u^{2} = 2 a s 0^{2} - 10^{2} = 2 (- 10) h h = 5 \frac{H}{2} = 5 H = 10 m$

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