Question

An arrow shot on s flat surface at an angle of 57° at 95m/s, have to find range, time in air, max height, height after 10 secs, and what is the velocity at those 10 secs

Answer 1

An arrow shot on a flat surface at an angle of 57° at 95m/s, have to find a range, time in air, max height, height after 10 secs, and what is the velocity at those 10 secs

Solution:

Let:

$$v = 95 \text{ m/s}$$ $$t = 10 \text{ sec}$$ $$\alpha = 57{}^\circ$$ $$S - ?, t_{\text{total}} - ?, H - ?, H_{(10)} - ?, v_{(10)} - ?$$$$S = v_x * t_{(\text{total})} = v t_{(\text{total})} \cos \alpha$$ $$t_{(\text{total})} = 2 \frac{v_y}{g} = 2 \frac{v \sin \alpha}{g}$$ $$t_{(\text{total})} = 2 * \frac{95 * \sin 57{}^\circ}{9.8} = 16.26 \text{ sec}$$ $$S = 95 * 16.26 * \cos 57{}^\circ = 841.3 \text{ m}$$ $$H = \frac{1}{2} g \left( \frac{1}{2} t_{(\text{total})} \right)^2 = \frac{1}{2} * 9.8 * (16.26)^2 = 323.88 \text{ m}$$

After 10 sec an arrow will be in free falling such as the maximum height is at $t = 1/2 t_{(\text{total})} = 8.13 \text{ sec}$

$$v_{(10)} = g \left(10 - \frac{1}{2} t_{(\text{total})}\right) = g * (10 - 8.13) = 18.33 \text{ m/sec}$$ $$H_{(10)} = H - \frac{1}{2} g \left(10 - \frac{1}{2} t_{(\text{total})}\right)^2 = 306.75 \text{ m}$$

**Answer:**

$$t_{(\text{total})} = 16.26 \text{ sec}, \quad S = 841.3 \text{ m}, \quad H = 323.88 \text{ m}, \quad v_{(10)} = 18.33 \frac{\text{m}}{\text{sec}}, \quad H_{(10)} = 306.75 \text{ m}$$