NEET Physics Motion in a Straight Line / सरल रेखा में गति NEET PYQ-One Dimensional Motion

A stone is thrown vertically upwards. When stone is at a height half of its maximum height, its speed is 10 m/s; men die maximum height attained by the stone is \[(g=10\text{ }m/{{s}^{2}})\]: [AIPMT 2001]

A) 8 m

B) 10 m

C) 15 m

D) 20 m

Correct Answer: B

Solution :

Let \[u\] be the initial velocity and H the maximum height attained.

When at height \[h=\frac{H}{2},\,\] we have

\[v={{v}_{1}}=10\text{ }m/s\]

From third equation of motion

\[v_{1}^{2}={{u}^{2}}-2gh\]

or \[{{(10)}^{2}}={{u}^{2}}-2g\frac{H}{2}\] ..(i)

At height H, \[{{v}_{2}}=0\]

\[v_{2}^{2}={{u}^{2}}-2gH\]

or \[0={{u}^{2}}-2gH\] ..(ii)

Subtract Eq. (ii) From Eq. (i), we get

\[{{(10)}^{2}}=2g\frac{H}{2}\]

or \[H=\frac{{{(10)}^{2}}}{g}\]

or \[H=\frac{{{(10)}^{2}}}{10}=10\,m\]

Alternative: maximum height attained by the stone

\[H=\frac{{{u}^{2}}}{2g}\]

When \[h=\frac{H}{2},\,u=10\,m/s\]

\[\frac{H}{2}=\frac{{{(10)}^{2}}}{2g}\]

or \[H=\frac{100}{10}=10\,m\]