Question

Two projectiles $\text{A}$ and $\text{B}$ are thrown with the same speed such that $\text{A}$ makes angle $\theta$ with the horizontal and $\text{B}$ makes angle $\theta$ with the vertical, then -

• A. Both must have same time of flight
• B. Both must achieve same maximum height
• C. $\text{A}$ must have more horizontal range than $\text{B}$
• D. Both may have same time of flight

Answer 1

The correct option is D Both may have same time of flight


From diagram,

$(u_x{)}_A=u\cos \theta ;(u_y{)}_A=u\sin \theta$

And, $(u_x{)}_B=u\sin \theta ;(u_y{)}_B=u\cos \theta$

Time of flight of a projectile motion is given by

$T=\frac{2u_y}{g}$

Since $(u_y{)}_A=u\sin \theta$ and $(u_y{)}_B=u\cos \theta$ are different.

So, $T_A$ and $T_B$ will be different when $\theta \ne {45}^{\circ }$ and will be same when $\theta ={45}^{\circ }$.

Hence, $T_A$ & $T_B$ may be same.

Similarly, for maximum height,

$h_{max}=\frac{u_y^2}{2g}$

again $(h_{max}{)}_A$ & $(h_{max}{)}_B$ must be different for $(\theta \ne {45}^{\circ })$

Horizontal Range of projectile motion is given by

$R=\frac{2}{g}{u}_x.u_y$

For $\text{A}$ and $\text{B}$, $u_x.u_y$ will be equal

So, $R_A=R_B$

From the above solution, we conclude that option $\left(d\right)$ is the correct choice.