Question

A projectile is fired at an upward angle of 45° from the top of a 233 m cliff with a speed of 192 m/s. What will be its speed when it strikes the ground below?

Answer 1

1. A projectile is fired at an upward angle of $45{}^{\circ}$ from the top of a 233 m cliff with a speed of 192 m/s. What will be its speed when it strikes the ground below?

Solution.

Given: $\theta = 45{}^{\circ}$, $h_0 = 939\,m$, $h = 233\,m$, $v = 192\,\frac{\mathrm{m}}{\mathrm{s}}$,

Find: $v_{1}$—?

$$v_{x} = \text{const} = v \cdot \cos 45{}^{\circ} = 192\,\frac{\mathrm{m}}{\mathrm{s}} \cdot \frac{\sqrt{2}}{2} = 135.765\,\frac{\mathrm{m}}{\mathrm{s}}$$ $$v_{y_{0}} = v \cdot \sin 45{}^{\circ} = 192\,\frac{\mathrm{m}}{\mathrm{s}} \cdot \frac{\sqrt{2}}{2} = 135.765\,\frac{\mathrm{m}}{\mathrm{s}}$$ $$h_{0} = \frac{v_{y_{0}}^{2}}{2g} = \frac{v_{y_{0}}^{2}}{2g} = 939\,m,$$ $$h_{0} + h = \frac{v_{y_{1}}^{2}}{2g}, \quad v_{y_{1}} = \sqrt{2g \cdot (h_{0} + h)} = \left(151.640\,\frac{\mathrm{m}}{\mathrm{s}}\right),$$ $$v_{1} = \sqrt{v_{y_{1}}^{2} + v_{y_{0}}^{2}} = \sqrt{\left(151.640\,\frac{\mathrm{m}}{\mathrm{s}}\right)^{2} + \left(135.765\,\frac{\mathrm{m}}{\mathrm{s}}\right)^{2}} = 203.536\,\frac{\mathrm{m}}{\mathrm{s}}$$

Answer:

$$v_{1} = 203.536\,\frac{\mathrm{m}}{\mathrm{s}}$$