Answer:10 sec
Explanation:
Given
Ball follows the path [tex]S(t)=-3t^2+24t+60[/tex]
Therefore at t=0
S(0)=60
i.e. height of building is 60 m
(a)Ball will hit when the distance between ball and ground is zero i.e.
[tex]0=-3t^2+24t+60[/tex]
[tex]3t^2-24t-60=0[/tex]
[tex]t^2-8t-20=0[/tex]
[tex]t^2-10t+2t-20=0[/tex]
[tex]\left ( t+2\right )\left ( t-10\right )=0[/tex]
i.e. at t=10 sec ball will hit ground
(b)accleration at time t=1 sec
acceleration is given by [tex]\frac{\mathrm{d^2} S(t)}{\mathrm{d} t^2}=\frac{\mathrm{d^2} \left ( -3t^2+24t+60\right )}{\mathrm{d} t^2}[/tex]
[tex]a=-6 m/s^2[/tex]
