InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A stone is allowed to fall form the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s calculate. When and where the two stones will meet. |
|
Answer» Solution :Let t be the point at which two stones meet and let H be their height from the ground. It is given in the equation that height of the tower is h=100m. Now first consider the STONE which falls from the top of the tower. So distance covered by this stone at time .t. can be calculated using equation of motion. `X-X_(o)=u_(o)t+3//2g"t^(2)` Since initial velocity U=o so we GET `100-x=1//2g""t^(2)""(1)` The distance covered by the same that is thrown in upwards direction from ground is `x=25t-1//2g""t^(2)""(2)` Adding equation 1 and 2 we get t=4s Putting value to eqn (2) `X=25xx4-1//2xx9.8xx(4)^(2)=100-78.4` =21.6m |
|