Question

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

The general equation is

$$y (t) = v _ {0} t - \frac {g t ^ {2}}{2}$$ $$t _ {1} = 2 s; t _ {2} = 2 s$$ $$y (t _ {1}) = v _ {0} t _ {1} - \frac {g t _ {1} ^ {2}}{2}$$ $$y (t _ {2}) = v _ {0} t _ {2} - \frac {g t _ {2} ^ {2}}{2}$$

We express the initial velocity from the two equations

$$y (t _ {1}) = v _ {0} t _ {1} - \frac {g t _ {1} ^ {2}}{2} \rightarrow v _ {0} = \frac {y (t _ {1}) + \frac {g t _ {1} ^ {2}}{2}}{t _ {1}}$$ $$y (t _ {2}) = v _ {0} t _ {2} - \frac {g t _ {2} ^ {2}}{2} \rightarrow v _ {0} = \frac {y (t _ {2}) + \frac {g t _ {2} ^ {2}}{2}}{t _ {2}}$$ $$- > \frac {y (t _ {1}) + \frac {g t _ {1} ^ {2}}{2}}{t _ {1}} = \frac {y (t _ {2}) + \frac {g t _ {2} ^ {2}}{2}}{t _ {2}}$$

Considering

$$y (t _ {1}) = y (t _ {2}) = h$$

We have $h = \frac{g(t_2^2t_1 - t_1^2t_2)}{2(t_2 - t_1)} = \frac{9.81(64*2 - 8*4)}{2(8 - 2)} = 78.48 \, \text{m}$