The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20ms-1 its time of flight is (g=10ms-2 )

Question

The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20ms-1 its time of flight is (g=10ms-2 ) (A) 2s (B) 2√2s (C) (1/2) s (D) (1/√2)s

Tardigrade Admin · Accepted Answer

P.E = K.E (1/2) mu2 sin2 θ = (1/2) mu2 cos2 θ then θ = 45° and use T= (2U sin θ/g)

Q. The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is $2 0_{m s^{- 1}}$ its time of flight is $(g = 1 0_{m s^{- 2}})$

2s

$22 s$

$\frac{1}{2} s$

$\frac{1}{2} s$

$P . E = K . E$
$\frac{1}{2} m u^{2} s i n^{2} θ = \frac{1}{2} m u^{2} co s^{2} θ$
then $θ = 4 5^{\circ}$ and use $T = \frac{2 U s in θ}{g}$

Q. The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is 20ms−1​ its time of flight is (g=10ms−2​)

2s

22​s

21​s

2​1​s

P.E=K.E 21​mu2sin2θ=21​mu2cos2θ then θ=45∘ and use T=g2Usinθ​

Q. The potential energy of a projectile at its maximum height is equal to its kinetic energy there. If the velocity of projection is $2 0_{m s^{- 1}}$ its time of flight is $(g = 1 0_{m s^{- 2}})$

$22 s$

$\frac{1}{2} s$

$\frac{1}{2} s$

$P . E = K . E$
$\frac{1}{2} m u^{2} s i n^{2} θ = \frac{1}{2} m u^{2} co s^{2} θ$
then $θ = 4 5^{\circ}$ and use $T = \frac{2 U s in θ}{g}$