Answer to Question #247757 in Mechanics | Relativity for ara

Question #247757

A rock is thrown horizontally from the top of a building with an initial speed of v = 10.1 m/s. If it lands d = 57.1 m from the base of the building, how high is the building?


1
Expert's answer
2021-10-06T15:52:13-0400

The speed in horizontal direction remains unchanged during the falling. Thus, the time of falling is given as follows:


t=dvht = \dfrac{d}{v_h}

where d=57.1md = 57.1m is the range, and vh=10.1m/sv_h = 10.1m/s is the horizontal speed.

The distance travelled in vertical direction (height of the building) is given by the following kinematic equation:


h=gt22h = \dfrac{gt^2}{2}

where g=9.81m/s2g = 9.81m/s^2 is the gravitational acceleration. Substituting the expression for tt, obtain:


h=gd22vh2=9.8157.12210.12157mh = \dfrac{gd^2}{2v_h^2} = \dfrac{9.81\cdot 57.1^2}{2\cdot 10.1^2} \approx 157m

Answer. 157 m.


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