Answer to Question #255857 in Mechanics | Relativity for Yenny

Question #255857

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



1
Expert's answer
2021-10-24T18:29:28-0400

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


h=gt22t=2hgh = \dfrac{gt^2}{2}\\ t = \sqrt{\dfrac{2h}{g}}

where h=18mh = 18m and g=9.8m/s2g = 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=u2hgd = ut = u\sqrt{\dfrac{2h}{g}}

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


d=252189.892md = 25\cdot \sqrt{\dfrac{2\cdot 18}{9.8}} \approx 92m

Answer. 92m.


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