Question

A body is thrown vertically upward such that it crosses the same height after 2 sec. andafter 8 sec.
What is the value of the mentioned height?

Answer 1

A body is thrown vertically upward such that it crosses the same height after 2 sec. and after 8 sec. What is the value of the mentioned height?

mentioned height equals:

$$h = \frac {v _ {0} ^ {2} - v _ {1} ^ {2}}{2 g}$$

where, $v_{0}$ – initial speed, $v_{1}$ – speed after 2 sec.

$v_{1}$ equals:

$$v _ {1} = v _ {0} - g * t _ {0}$$

where, $t_0 = 2$ sec

Therefore: $v_{0} = v_{1} + 2g$

In another side:

$$2 v _ {1} = g * (t _ {1} - t _ {0})$$

where $t_1 = 8$ sec, therefore $v_1 = 3g$

$$v _ {0} = v _ {1} + 2 g = 3 g + 2 g = 5 g$$

From first equation:

$$h = \frac {v _ {0} ^ {2} - v _ {1} ^ {2}}{2 g} = g * \frac {5 ^ {2} - 3 ^ {2}}{2} = 80 m$$

Answer: 80 m