Trajectories of two projectiles are shown in figure. Let T1 and T2 be the time of flight and u1 and u2 are their speeds of projections, then

Question

(A) T2 > T1 (B) T2 = T1 (C) u1 > u2 (D) u1 < u2

Tardigrade Admin · Accepted Answer

Since the height of both the projectiles are same, we have: H 1= H 2 ( u 12 sin 2 θ1/2 g )=( u 22 sin 2 θ2/2 g ) u 12 sin 2 θ1= u 22 sin 2 θ2 u 1 sin θ1= u 2 sin θ2 wher u is initial speed and θ is the angle of projectile. ( T 1/ T 2)=((2 u 1 sin θ1/ g )/(2 u 2 sin θ2) g ) =(2 u 1 sin θ1/ g ) × ( g /2 u 2 sin θ2) =( u 1 sin θ1/ u 2 sin θ2)=1 T 1= T 2

Q. Trajectories of two projectiles are shown in figure. Let $T_{1}$ and $T_{2}$ be the time of flight and $u_{1}$ and $u_{2}$ are their speeds of projections, then

$T_{2} > T_{1}$

$T_{2} = T_{1}$

$u_{1} > u_{2}$

$u_{1} < u_{2}$

Q. Trajectories of two projectiles are shown in figure. Let T1​ and T2​ be the time of flight and u1​ and u2​ are their speeds of projections, then

T2​>T1​

T2​=T1​

u1​>u2​

u1​<u2​

Q. Trajectories of two projectiles are shown in figure. Let $T_{1}$ and $T_{2}$ be the time of flight and $u_{1}$ and $u_{2}$ are their speeds of projections, then

$T_{2} > T_{1}$

$T_{2} = T_{1}$

$u_{1} > u_{2}$

$u_{1} < u_{2}$