Two projectiles A and B are thrown with the s | vedclass

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(D) Let the speed of projection be $u$. For projectile $A$,the angle with the horizontal is $	heta_A = 	heta$. For projectile $B$,the angle with the

Vedclass · Accepted Answer

Both may have the same time of flight.

Vedclass · Answer

(D) Let the speed of projection be $u$. For projectile $A$,the angle with the horizontal is $	heta_A = 	heta$. For projectile $B$,the angle with the vertical is $	heta$,so the angle with the horizontal is $	heta_B = 90^{\circ} - 	heta$.Since $	heta_B$ is the complement of $	heta_A$,these are complementary angles.For complementary angles,the horizontal range is the same $(R_A = R_B)$.However,the time of flight $T = \frac{2u \sin 	heta}{g}$ and maximum height $H = \frac{u^2 \sin^2 	heta}{2g}$ depend on the specific angle $	heta$.If $	heta = 45^{\circ}$,then $	heta_A = 	heta_B = 45^{\circ}$,and all parameters (time of flight,maximum height,and range) will be identical.If $	heta 
eq 45^{\circ}$,the time of flight and maximum height will be different for $A$ and $B$.Therefore,the statement that they 'may' have the same time of flight is correct,as it holds true for the case $	heta = 45^{\circ}$.

Two projectiles $A$ and $B$ are thrown with the same speed such that $A$ makes an angle $\theta$ with the horizontal and $B$ makes an angle $\theta$ with the vertical. Then:

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