An object is launched at 64 feet per second from an 80-foot tall platform. The equation for the object's height #s# at time #t# seconds after launch is #s(t) = -16t^2 + 64t + 80#, where #s# is in feet. When does the object strike the ground?

1 Answer
P dilip_k
May 14, 2016

5 sec

Explanation:

The equation of object’s height s at time t sec after launch from 80 feet tall platform is
#s(t)-16t^2+64 t +80……(1)#
We are to determine the time of reaching ground after launch.
At t= 0 the height s(0)=80ft,inserting t =0 in equation(1)

When it will reach the ground s(t) will be zero, So putting s(t)=0 in equation (1) we have
#0=-16t^2+64t+80#
#=>-16t^2+64t+80=0#
#=>t^2-4t-5=0#
#=>t^2-5t+t-5=0#
#=>t(t-5)+1(t-5)=0#
#=>(t-5)(t+1)=0#
Neglecting negative value of t , the required time of reaching ground is t = 5 sec

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