A stone is dropped from the top of a tower of height `h`. Aftre `1 s` – InterviewSolution

InterviewSolution

1.	A stone is dropped from the top of a tower of height `h`. Aftre `1 s` another stone is droppped from the balcony `20 m` below the top. Both reach the bottom simultaneously. What is the value of `h` ? Take `g=10 ms^(-2)`.A. 43125mB. 312.5mC. 31.25mD. 25.31m
Answer» Correct Answer - C `h=(1)/(2)"gt"^(2)` and `h-20=(1)/(2)g(t-1)^(2)` `(1)/(2)"gt"^(2)-20=(1)/(2)"gt"^(2)-20=(1)/(2)"gt"^(2)-gt+(g)/(2)` `gt=20+(g)/(2)` `10t=20+5rArr t=2.5` `h=(1)/(2)"gt"^(2)=(1)/(2)xx10xx(2.5)^(2)` `=5xx6.25=31.25m`

Discussion

No Comment Found

Related InterviewSolutions

In the above question, the acceleration of the rocket is :A. `50ms^(-2)`B. `100ms^(-2)`C. `10 ms^(-2)`D. `1000ms^(-2)`
In above question the retardation of rocket is:-A. `50ms^(-2)`B. `100ms^(-2)`C. `500ms^(-2)`D. `10ms^(-2)`
In the previous question, the mean velocity of the rocket reaching the maximum height is :A. `100ms^(-1)`B. `50ms^(-1)`C. `500ms^(-1)`D. `25//3ms^(-1)`
In the previous question, the height attained by the rocket before deceleration is :A. 1 kmB. 10kmC. 20kmD. 60 km
A ball is projected to attain the maximum range. If the height attained is H, the range isA. HB. 2HC. 4HD. H/2
In the Q.23, the maximum height attained by the arrow is:-A. 25mB. 40mC. 31.25mD. 12.5m
A body stans at `78.4 m` from a building and throws a ball which just enters a window `39.2 m` above the ground. Calculate the velocity of projection of the ball. Fig. 2 (d) . 22. .
The equation of projectile is `y=16x-(x^(2))/(4)` the horizontal range is:-A. 16mB. 8mC. 64mD. 12.8m
If R is the maximum horizontal range of a particle, then the greatest height attained by it is :A. RB. 2RC. `(R)/(2)`D. `(R)/(4)`
What is the ratio of P.E. w.r.t. ground and K.E. at the top most point of the projectile motion:-A. `cos^(2)theta`B. `sin^(2)theta`C. `tan^(2)theta`D. `cot^(2)theta`