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>
Step-by-step explanation:
• y2 → -1
• y1 → 5
• x1 → 4
• x2 → -3
I think the answer is -48°F
Okay. Handle 500-100 first. That brings it to 3{5[10+5(400)+399]}. Next, Handle 5(400). That brings you to 3{5{10+2,000+399]}. Then handle all the addition. That leads to 3[5(2,409)]. Next, configure 5(2409). That leads to 3(12045). Then do that one which ends as 36,135 for a final answer.
