A ball is thrown vertically upward. It has a speed of 10 m/s when it has reached one half of its maximum height. How high does the ball rise? \[(Taking\,\,g=10m/{{s}^{2}})\]
A)15 m
B)10 m
C)20 m
D)5 m
Correct Answer:
B
Solution :
Let maximum height attained by the ball be H. Using \[{{v}^{2}}={{u}^{2}}-2gh\] \[At\,\,A,\,\,{{(10)}^{2}}\,=\,\,{{u}^{2}}-2\times 10\times \frac{H}{2}\] \[\Rightarrow \,\,\,{{u}^{2}}=100\,\,+\,\,10\,H\] ? (i) \[At\text{ }O',\text{ }{{(0)}^{2}}={{u}^{2}}-2\times 10\times H\] \[\Rightarrow \,\,\,{{u}^{2}}= 20\,\,H\] Thus, from Eqs. (i) and (ii), we get \[20\,\,H=100+10\,\text{H}\] \[\Rightarrow \,\,\,10\,\,H=100\,\,\,\,\,\,\therefore \,\,\,H=10\,m\]