Answer:
a) [tex]H=25.1020\ m[/tex]
b) [tex]t=3.2838\ s[/tex]
Explanation:
Given:
a)
Maximum height attained by the ball above the roof level can be given by the equation of motion.
As,
[tex]v^2=u^2-2g.h'[/tex]
where:
[tex]v=[/tex] final velocity at the top height of the upward motion [tex]=0\ m.s^{-1}[/tex]
[tex]g=[/tex] acceleration due to gravity
[tex]h'=[/tex] height of the ball above the roof
Now,
[tex]0^2=10^2-2\times 9.8\times h'[/tex]
[tex]h'=5.10\ m[/tex]
Therefore total height above the ground:
[tex]H=h+h'[/tex]
[tex]H=20+5.1020[/tex]
[tex]H=25.1020\ m[/tex]
b)
Now we find the time taken in raching the height [tex]h'[/tex]:
[tex]v=u-gt'[/tex]
[tex]v=[/tex] final velocity at the top of the motion [tex]=0\ m.s^{-1}[/tex]
So,
[tex]0=10-9.8\times t'[/tex]
[tex]t'=1.0204\ s[/tex]
Now the time taken in coming down to the ground from the top height:
[tex]H=u'.t_d+\frac{1}{2} g.t_d^2[/tex]
where:
[tex]u'=[/tex] is the initial velocity of the ball in course of coming down to ground from the top [tex]=0\ m.s^{-1}[/tex]
Here the direction acceleration due to gravity is same as that of motion so we are taking them positively.
[tex]25.1020=0+0.5\times 9.8\times t_d^2[/tex]
[tex]t_d=2.2634\ s[/tex]
Therefore the total time taken in by the ball to hit the ground after it begins its motion:
[tex]t=t'+t_d[/tex]
[tex]t=1.0204+2.2634[/tex]
[tex]t=3.2838\ s[/tex]
