Question

You throw a ball straight up with an initial velocity of 14.3 m/s. On the way up it passes a tree branch at a height of 7.8 m. How much additional time will pass before the ball passes the tree branch on the way back down?
Numeric: A numeric value is expected and not an expression.
t = __________________________________________

Answer 1

Answer on Question#60306 -Physics- Mechanics -Relativity

You throw a ball straight up with an initial velocity of $14.3\mathrm{m / s}$ . On the way up it passes a tree branch at a height of $7.8\mathrm{m}$ . How much additional time will pass before the ball passes the tree branch on the way back down? Numeric: A numeric value is expected and not an expression.

t =

Solution.

$v_{0} = 14.3\mathrm{m / s}$

$g = 9.8\mathrm{m / s}^2$

$h = 7.8\mathrm{m}$

$\Delta t - ?$

The movement of the ball consists of two parts. First, the ball moves upward until its speed reaches zero then moves down. The ball moves with an initial velocity $v_{0} = 14.3 \, \text{m/s}$ and constant acceleration $g = 9.8 \, \text{m/s}^{2}$ downwards. Hence the height of the ball is described by the equation

$$\begin{array}{l} h = v _ {0} t - \frac {g t ^ {2}}{2} \rightarrow 7. 8 = 1 4. 3 t - 4. 9 t ^ {2} \\ 4. 9 t ^ {2} - 1 4. 3 t + 7. 8 = 0 \\ \end{array}$$

Solve this quadratic equation

$$D = 1 4. 3 ^ {2} - 4 \cdot 4. 9 \cdot 7. 8 = 5 1. 6 1$$

$t_1 = \frac{14.3 - \sqrt{51.61}}{2 \cdot 4.9} \approx 0.7$ s and $t_2 = \frac{14.3 + \sqrt{51.61}}{2 \cdot 4.9} \approx 2.2$ s

$t_1$ corresponds to the time during ascent to a height of $h$ after the start of movement; the time $t_2$ corresponds to the time of descent to a height of $h$ after the start of movement. Therefore $\Delta t = t_2 - t_1 = 2.2 - 0.7 = 1.5$ s

Answer: $\Delta t = 1.5$ s

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