Question

Q2. On a cold day, hailstones fall with a velocity of (2i− 6k) m s−1 . If a cyclist travels through the hail at 10i ms−1 , what is the velocity of the hail relative to the cyclist? At what angle are the hailstones falling relative to the cyclist

Answer 1

Answer:$-8\hat{i}-6\hat{k}$

$\theta =\tan^{-1}\left ( \frac{3}{4} \right )$

Step-by-step explanation:

Given

Velocity of hailstones fall$\left ( V_h\right )=2\hat{i}-6\hat{k}$ m/s

Velocity of cyclist $\left ( V_c\right )=10\hat{i}$ m/s

Therefore

Velocity of hail with respect to cyclist$\left ( V_{hc}\right )$

$V_{hc}=V_h-V_c$

$V_{hc}=2\hat{i}-6\hat{k}-10\hat{i}$

$V_{hc}=-8\hat{i}-6\hat{k}$

and angle of hails falling relative to the cyclist is given by

$\theta =\tan^{-1}\left ( \frac{3}{4}\right )$

$\theta$ is the angle made with the vertical