Question #7cc65

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A08
Jan 28, 2018

(2)

Explanation:

Applicable kinematic expression is

#h=ut+1/2g t^2# .....(1)

Lets choose origin of the coordinate system at the point of projection of object. When object reaches ground level, #h=-58.8m#. taking value of #g=9.8ms^-2# and inserting values in (1) we get

#-58.8=4.9t-1/2xx9.8xxt^2#
#=>4.9t^2-4.9t-58.8=0#

Diving both sides with #4.9# equation becomes

#=>t^2-t-12=0#

The quadratic can be solved by splitting the middle term.

#=>t^2-4t+3t-12=0#
#=>t(t-4)+3(t-4)=0#
#=>(t-4)(t+3)=0#

Setting both factors equal to zero, we get two roots as #t=4# and #t=-3#. Ignoring the #-ve# root as time can not be #-ve#, we get

#t=4.0s#

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