Saved Bookmarks
| 1. |
A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown( one upwards and one downwards ) with the same velocity u and they reach the earth surface after t1 and t3 seconds respectively , then1) t= t1 - t2 2) t = t1+ t2 ÷23) t=√t1t2 |
|
Answer» case 1 : A stone is DROPPED from a building of height h and time taken by stone to reach the ground is t. then, h = 1/2 gt² ......(1) case 2 : same stone is thrown VERTICALLY downward with SPEED u m/s. time taken by stone to reach the ground is t1. so, -h = -ut1 + 1/2 (-g)t1² h = ut1 + 1/2 gt1² gt1² + 2ut1 -2h = 0 .......(2) t1 = {-u + √(u² +2gh)}/g [ t1 ≠ {-u-√(u²+2gh)}/g] case 3 : same stone is thrown vertically upwards with speed u m/s and time taken by stone to reach the ground is t2. so, -h = ut2 + 1/2 (-g)t2² or, h = -ut2 + 1/2 gt2² or, gt2² - 2ut2 - 2h = 0.......(3) t2 = {u + √(u² + 2gh)}/g [ t2 ≠ {u - √(u² + 2gh)}/g] now, t1.t2 =[ {-u + √(u² +2gh)}/g ]. [ {u + √(u² +2gh)}/g ] = (u² + 2gh - u²)/g² = 2h/g from equation (1), t1.t2 = 2h/g = t² or, t = √(t1.t2) This is required answer. |
|