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 ( (3)/(4) \right )`

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 ( (3)/(4)\right )`

`\theta` is the angle made with the vertical