Question

A car is driving down the road at a velocity $V_c$ relative to ground and is delivering newspapers to homes. The newspapers are thrown at a velocity of $V_p$ relative to the car. At what angle $\theta$ must the newspaper be thrown, relative to the car, so that they fly in a direction parallel to the driveways?

• A. $\theta ={\sin }^{-1}\left(V_c/V_p\right)$
• B. $\theta ={\sin }^{-1}\left(V_p/V_c\right)$
• C. $\theta ={\cos }^{-1}\left(V_c/V_p\right)$
• D. $\theta ={\cos }^{-1}\left(V_p/V_c\right)$

Answer 1

The correct option is A $\theta ={\sin }^{-1}\left(V_c/V_p\right)$

For the newspapers to fly parallel to the driveways when they are airborne, the component of velocity of $V_p$ that is parallel to the road must cancel out with $V_c$, so that the velocity of the newspaper with respect to ground is in a direction parallel to the driveways.

$V_{car}=V_c\hat{i}$
$V_{\text{newspaper,car}}=V_P\cos \theta \hat{j}-V_P\sin \theta \hat{i}$
Let the velocity of newspaper$V_N$ w.r.t ground is in the direction of the driveway.
$V_{\text{newspaper,car}}=V_N-V_{car}$
$V_P\cos \theta \hat{j}-V_P\sin \theta \hat{i}=V_N\hat{j}-V_c\hat{i}$
The component along the road must be zero.
Therefore, $V_P\sin \theta =V_c$
$\Rightarrow \theta ={\sin }^{-1}\left(V_c/V_p\right)$