Answer to Question #255857 in Mechanics | Relativity for Yenny

Question

What is the range of the projectile launched horizontally at 25 m/s from the 18 m-high cliff edge

Answer 1

Since horizontal and vertical motions are independent, the time of fall can be found from the following kinematic equation:

$$h = \dfrac{gt^2}{2}\\ t = \sqrt{\dfrac{2h}{g}}$$

where $h = 18m$ and $g = 9.8m/s^2$.

For the same reason the initial speed in horizontal direction does not change, and the distance travelled in this direction (range) can be found as follows:

$$d = ut = u\sqrt{\dfrac{2h}{g}}$$

where $u = 25m/s$. Thus, obtain:

$$d = 25\cdot \sqrt{\dfrac{2\cdot 18}{9.8}} \approx 92m$$

Answer. 92m.