Question

An object thrown vertically up from the ground passes the height 5 m twice in an interval of 10 s.What is its time of flight?

Answer 1

An object thrown vertically up from the ground passes the height 5 m twice in an interval of 10 s. What is its time of flight?

**Soltuion:**

H = 5m

t = 10s

T – time of the flight

If the ball flies twice a height of 5 meters in 10 seconds, then on top of it was $t_2 = 5$ seconds after the first span along the height of 5m.

Rate equation for the ball, in the end of the flight speed of the ball is zero:

$$0 = V_1 - g t_2$$ $$V_1 = g t_2$$

The equation of motion for the ball for the height of 5 meters:

$$y: H = V t_1 - \frac{g t_1^2}{2} , \quad t_1 - \text{flight time to reach the height } H = 5\,\text{m}$$

Rate equation for the ball

$$V_1 = V - g t_1$$ $$V = V_1 + g t_1$$ $$(1)\text{in}(3): \quad V = g t_1 + g t_2$$ $$(4)\text{in}(2): \quad H = g (t_1 + t_2) t_1 - \frac{g t_1^2}{2}$$ $$g t_1^2 + 2 g t_2 t_1 - 2H = 0$$

We have a quadratic equation:

$$9.8 t_1^2 - 98 t_1 - 10 = 0$$

Real positive root of the equation: $t_1 = 10.1\,\text{s}$

Flight time up is equal time of the fall:

$$T = 2 \cdot (t_1 + t_2) = 2 \cdot (10.1\,\text{s} + 5\,\text{s}) = 30.2\,\text{s}$$

Answer: time of the flight is 30.2s