<u>We are given:</u>
The function: y = -16t² + 64
where y is the height from ground, t seconds after falling
<u>Part A:</u>
when the droplet would hit the ground, it's height from the ground will be 0
replacing that in the given function:
0 = -16t² + 64
16t² = 64 [adding 16t² on both sides]
t² = 4 [dividing both sides by 16]
t = 2 seconds [taking square root of both sides]
<u>Part B:</u>
for second droplet,
height from ground = 16 feet
time taken = t seconds
acceleration due to gravity = 10 m/s²
initial velocity = 0 m/s
h = ut + (1/2)at² [second equation of motion]
16 = (0)(t) + (1/2)(10)(t²)
16 = 5t²
t² = 16/5
t = 1.8 seconds (approx)
Therefore, the second droplet takes the least amount time to hit the ground
- Use the formula V = π <span>∙ r2 ∙ h.
- Plug into the equation. 27</span>π(cubed)=π∙3(2)∙h
-Solve. 27π(3)=9π∙h
27π/9π=3π
I believe the height is 3π.
<span>Hope this helps
</span>
Answer:
I believe it's "Multiplication Property of Equality"
The answer is 78 because you divide 2106 and 27 to find the length
