Question

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

Answer 1

Question #32443

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?

Solution:

Let:

$$t _ {1} = 2 \sec$$ $$t _ {2} = 8 \sec$$ $$h _ {1} = ?$$

For the body thrown vertically upwards

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

Were $v_{0}$ is the initial velocity $g$ is the acceleration due the gravity

Such as the Uprise time of the body from mentioned height to maximal height is equal to the slope time from the maximal height to mentioned height

$$t _ {\text {maximal height}} = \frac {t _ {2} - t _ {1}}{2} + t _ {1}$$ $$t _ {\text {maximal height}} = \frac {t _ {2} + t _ {1}}{2}$$

Such as the velocity in highest point is equal to zero

$$v _ {0} = g t _ {\text {maximal height}}$$ $$v _ {0} = g \frac {t _ {2} + t _ {1}}{2}$$

According this

$$h _ {1} = g \frac {t _ {2} + t _ {1}}{2} t _ {1} - \frac {1}{2} g t _ {1} ^ {2}$$ $$h _ {1} = 9. 8 \frac {8 + 2}{2} 2 - \frac {1}{2} 9. 8 * 2 ^ {2} = 7 8. 4 m$$

Answer: 78.4 m.