InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A body is projected with a speed of 40 m/s vertically up from the ground. What is the maximum height reached by the body? What is the entrie time of motion? What is the velocity at 5 seconds after the projection? Take g = 10 m/s2. |
|
Answer» Initial speed u = 40 m/s ; g = 10 m/s2 Maximum height reached (h) = \(\frac{u^2}{2\,g}=\frac{40\,\times\,40}{2\,\times\,10}\) = 80 m Entire time of motion (T) = \(\frac{2u}{g}=\frac{2\,\times\,40}{10}\) = 8 s Entire time of motion is 8 seconds. ∴ It starts to fall down after 4 seconds. At 5 seconds the body is in downward direction. u = 0 m/s, a = g = 10 m/s2, t = 5 – 4 = 1 sec. v = u + at = 0 + 10 × 1 = 10 m/s ∴ The velocity at 5 seconds is 10 m/s downward. |
|