A stone is thrown at an angle θ to the horizontal reaches a maximum height h. The time of flight of the stone is:

Question

A stone is thrown at an angle θ to the horizontal reaches a maximum height h. The time of flight of the stone is: (A)  √((2h sin θ /g)  (B)  2√((2h sin θ /g)  (C)  2√((2h)/g)  (D)  √((2h)/g)

Tardigrade Admin · Accepted Answer

Maximum height attained by projectile is h=(u2 sin θ /2g) ?(i) ∴ u sin θ =√2gh ?(ii) Time of flight is given by T=(2u sin θ /g) ?(iii) From Eqs. (ii) and (iii), we get the time of flight is =(2√2gh/g)=2√(2gh/g2)=2√(2h/g)

Q. A stone is thrown at an angle $θ$ to the horizontal reaches a maximum height h. The time of flight of the stone is:

$\frac{( 2 h s i n θ}{g}$

$2 \frac{( 2 h s i n θ}{g}$

$2 \frac{( 2 h )}{g}$

$\frac{( 2 h )}{g}$

Maximum height attained by projectile is $h = \frac{u ^{2} s i n θ}{2 g}$ ?(i) $∴$ $u sin θ = 2 g h$ ?(ii) Time of flight is given by $T = \frac{2 u s i n θ}{g}$ ?(iii) From Eqs. (ii) and (iii), we get the time of flight is $= \frac{2 2 g h}{g} = 2 \frac{2 g h}{g ^{2}} = 2 \frac{2 h}{g}$

Q. A stone is thrown at an angle θ to the horizontal reaches a maximum height h. The time of flight of the stone is:

g(2hsinθ​​

2g(2hsinθ​​

2g(2h)​​

g(2h)​​