kiribaku176
Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Help, please! This is due tomorrow evening! Show work, please! Will give brainliest!

Help Please This Is Due Tomorrow Evening Show Work Please Will Give Brainliest class=
Help Please This Is Due Tomorrow Evening Show Work Please Will Give Brainliest class=
Help Please This Is Due Tomorrow Evening Show Work Please Will Give Brainliest class=
Help Please This Is Due Tomorrow Evening Show Work Please Will Give Brainliest class=

Sagot :

Answer:

h = height reached

-16 t^2 = 1/2 g t^2 where g = 32 ft/sec^2

v t = height due to original vertical speed v

s = initial height

1. -45 = -16 t^2 + 62 t + 0         measuring height from original height

solving quadratic

2. t = 4.5 sec

3. time to reach max height = v / g = 62/32 = 1.94 sec

H (above release point) = -16 (1.94^2) + 62 * 1.94 = 60 ft

4. 2 * 1.94 = 3.88    solved in part 3

5. 1.94   solved on part 3

Check:  time to fall 60 + 45 ft = 105 ft

105 = 1/2 g t^2

t = (210 / 32)^1/2 = 2.56 sec

total time = 2.56 * 1.94 = 4.5 sec     time to fall + rise time

-45 = -16 * 4.5^2 + 62 * 4.5

-45 = -324 + 279 = -45        checking time to reach -45 feet using given equation

scunningham084

Answer:

p

Step-by-step explanation:

Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.