Question

A projectile is fired at an upward angle of 35.0 degrees from the top of a 225-m cliff with a speed of 235m/s . What will be its speed when it strikes the ground below? (Use conservation of energy.)

Answer 1

1. A projectile is fired at an upward angle of 35.0 degrees from the top of a 225-m cliff with a speed of $235\mathrm{m / s}$ . What will be its speed when it strikes the ground below? (Use conservation of energy.)

Let check the dimension.

$$\left[ v _ {1} \right] = \sqrt {\left( \begin{array}{l} m \\ s \end{array} \right) ^ {2} + \frac {m}{s ^ {2}} \cdot m} = \sqrt {\frac {m ^ {2}}{s ^ {2}} = \frac {m}{s}}.$$

Let evaluate the quantity.

$$v _ {1} = \sqrt {2 3 5 ^ {2} + 2 \cdot 9 . 8 \cdot 2 2 5} \approx 2 4 4 \left( \begin{array}{c} m \\ s \end{array} \right).$$

Answer: $224 \frac{m}{s}$ .