InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A projectile is thrown in the upward direction making an angle of `60^@` with the horizontal direction with a velocity of `150 ms^-1`. Then the time after which its inclination with the horizontal is `45^@` isA. `15 (sqrt(3) - 1) s`B. `15 (sqrt(3) + 1) s`C. `7.5 (sqrt(3) - 1) s`D. `7.5 (sqrt(3) + 1) s` |
|
Answer» Correct Answer - C ( c) At the two points of the trajectory during projectile motion, the horizontal component of the velocity is same. Then `150 xx (1)/(2) = v xx (1)/(sqrt(2))` or `v = (150)/(sqrt(2)) ms^-1` Initially : `u_y = u sin 60^@ = (150 sqrt(3))/(2) ms^-1` Finally : `v_y = v sin 45^@ = (150)/(sqrt(2)) xx (1)/(sqrt(2)) = (150)/(2) ms^-1` But `v_y = u_y + a_y t` or `(150)/(2) = (150 sqrt(3))/(2) - 10 t` `10 t = (150)/(2) (sqrt(3) - 1)` or `t = 7.5 (sqrt(3)- 1)`. |
|