- Home
- Standard 11
- Physics
2.Motion in Straight Line
hard
A ball is thrown vertically upwards with a velocity of $19.6\,ms ^{-1}$ from the top of a tower. The ball strikes the ground after $6\,s$. The height from the ground up to which the ball can rise will be $\left(\frac{ k }{5}\right) m .$ The value of $k$ is ..... (use $g =9.8\,m / s ^{2}$)
A
$393$
B
$390$
C
$392$
D
$391$
(JEE MAIN-2022)
Solution
$t _{ a }=\frac{ u }{ g }=\frac{19.6}{9.8}=2 s$
$t _{ d }=6-2 s =\sqrt{\frac{2 h _{\max }}{ g }}$
$\Rightarrow h _{\max }=\frac{16 \times 9.8}{2}=\frac{392}{5}$
Standard 11
Physics
Similar Questions
easy