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>
<h3>
Answer: Linear</h3>
Each time x goes up by 3, y goes up by 2. The rate of change is constant.
slope = (change in y)/(change in x) = 2/3
y intercept = 6 since this is where x = 0
The equation of this line is y = (2/3)x+6
Answer:
12.5
Step-by-step explanation:
30-60-90 triangles always have the height be sqrt3 the base.
Answer:
Awsten Knight
Step-by-step explanation:
