Answer:
a)t = 0.323 sec
b)t'= 0.323 sec
c)|h|= 0.52 m
d)y=1.56 m
Explanation:
The speed of ball ,U=104 km/h
U = 104 x 5/18 m/s =28.88 m/s
x= 18.3 m
a)
The half of distance = 18.3/2=9.15 m
We know that
Distance = Velocity x time
9.15 = 28.28 .t
t = 0.323 sec
b)
The total time to cover distance 18.3 m
18.3 = 28.88 x T
T=0.646 m
So the second half distance will cover in t' time
t' = T -t
t' = 0.646 - 0.323
t' = 0.323 s
c)
Initial velocity in y direction is zero.
Vo= 0
[tex]h=V_o.t-\dfrac{1}{2}gt^2[/tex]
[tex]h=0-\dfrac{1}{2}\ttimes 10\times 0.323^2[/tex]
h= - 0.52 m ( this is displacement and we know that it is a vector)
So magnitude of |h|= 0.52 m
d)
[tex]H=V_o.T-\dfrac{1}{2}gT^2[/tex]
[tex]H=0-\dfrac{1}{2}\times 10\times 0.646^2[/tex]
H= - 2.08 m
|H|= 2.08 m
So the in second half y = H- h
y= 2.08 - 0.52 m
y=1.56 m
