A.)
<span>s= 30m
u = ? ( initial velocity of the object )
a = 9.81 m/s^2 ( accn of free fall )
t = 1.5 s
s = ut + 1/2 at^2
\[u = \frac{ S - 1/2 a t^2 }{ t }\]
\[u = \frac{ 30 - ( 0.5 \times 9.81 \times 1.5^2) }{ 1.5 } \]
\[u = 12.6 m/s\]
</span>
b.)
<span>s = ut + 1/2 a t^2
u = 0 ,
s = 1/2 a t^2
\[s = \frac{ 1 }{ 2 } \times a \times t ^{2}\]
\[s = \frac{ 1 }{ 2 } \times 9.81 \times \left( \frac{ 12.6 }{ 9.81 } \right)^{2}\]
\[s = 8.0917...\]
\[therfore total distance = 8.0917 + 30 = 38.0917.. = 38.1 m \] </span>
(x/(2x))+((2*3)/(2x))=3/4
(x+6)/2x=3/4
4x+24=6x
24=2x
x=12
Answer:
the answer would be 4 Miles.
explanation:
We already know that 5,280 feet equals 1 Mile, and so we divide-
21,120 divided by 5,280, and that equals 4.
hope that made sense! I’m not very good with wording things! :)
Answer:
x-1
Hope it helps
Iam sure about the answer
