Question

A ball is dropped from the top of a tower, after 2 s another ball is thrown vertically downwards with a speed of 40 m/s. After how much time and what distance below the top of tower the ball meets? ​

Answer 1

Solution

the first one takes greater time of travel than the second one ...

the time is greater by 2 sec...than the second one...

if the second one travels for t seconds before meeting the first one then...

then the first one will travel for (t+2) seconds

distance traveled by them are the same so....

$\frac{1}{2} g(t + 2) {}^{2} = ut + \frac{1}{2} g(t) {}^{2} \\ = > \frac{1}{2} \times 10 \times ( t + 2){}^{2} = 40t + \frac{1}{2} \times 10 \times (t) {}^{2} \\ = > 5(t +2 ) {}^{2} = 40t + 5t {}^{2} \\ = > 20t + 20 = 40t \\ = > t = 1 \: \: sec$

now...distance traveled by the ball

$= \frac{1}{2} \times 10 \times (1 + 2) {}^{2} \\ = 5 \times 9 \\ = 45 \: \: metre \\ \: \: from \: the \: top \: of \: the \: tower$

Answer 2

Explanation:

Steps for solving this one :

1. we have to get equations of Height i.e. h=1/2gt^2 & h=40(t-2)1/2×g(t-2)^2
2. we have to equate both of the equations
3. put 40(t-2) in LHS and 1/2gt^2 in RHS
4. get 1/2g common
5. cancel the t^2 - t^2
6. and find the value of time
7. then put the value of time in equation 1 i.e. 1/2 gt^2
8. and we can get value of height