Question #290376

A cannonball is fired horizontally at 23.40 m/s from the edge of a 55 m building. Determine the: (a) height of the cannonball from the ground after 2 s. (b) time taken by the cannonball to reach the ground. (c) distance from the base of the building to the point where the its landed.


1
Expert's answer
2022-01-24T13:15:57-0500

Given:

v=23.40m/sv=\rm 23.40\: m/s

H=55mH=\rm55\: m

g=9.8m/s2g=\rm 9.8\:m/s^2


(a) height of the cannonball from the ground after 2 s

h=Hgt2/2=559.822/2=35.4mh=H-gt^2/2=55-9.8*2^2/2=35.4\:\rm m

(b) time taken by the cannonball to reach the ground

t=2Hg=2559.8=3.35st=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2*55}{9.8}}=3.35\: \rm s

(c) distance from the base of the building to the point where the its landed

L=vt=23.403.35=78.4  mL=vt=23.40*3.35=78.4\;\rm m


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