Question

9. An object is dropped from the top of the tower of height 80 m. One second later, another object
is thrown downwards with some velocity. The two objects reach the ground simultaneously
the velocity with which the second object was thrown ?​

Answer 1

Answer:

Time taken by first object

$= \sqrt{2h \over \: g} = \sqrt{ \frac{2 \times 80}{10} } = 4 \: s$

Let velocity be u of second particle . After

4 - 1 =3s it travelled 80 m .

$s = ut + \frac{1}{2} g {t}^{2} \\ \\ 80 = u(3) + \frac{1}{2} \times 10 \times {3}^{2} \\ \\ 3u = 80 - 45 = 35 \\ \\ u = 11.67 \: m {s}^{ - 1}$