Question

last one
@BlazeRyder
5. Suppose a skydiver (mass = 75 kg) is falling toward the Earth. When the skydiver is 100 m above the Earth he is moving at 60 m/s. At this point calculate the skydiver’s…….
a. (2 pts.) Gravitational potential energy





b. (2 pts.) Kinetic energy





c. (2 pts.) Total mechanical energy (The sum of kinetic and potential energy)

Answer 1

Alright, so it would be the same 2 formulas: \[P.E.=\frac{ 1 }{ 2 } \times m \times v ^{2}\] and \[P.E._{g}=m \times g \times h\] So from the question: m=75kg H=100m, and v=60m/s So tell me what you get for A and B when we can work on C.

Answer 2

*then we can work on c

Answer 3

i don't know

Answer 4

\[PE=0.5 \times 75 \times 60^{2}\]

\[PE _{g}=75 \times 9.8 \times 100\] put those into a calculator and tell me what you get.

Answer 5

for the first i think it's 2250

Answer 6

the second one is 73500

Answer 7

i think that's what the second one is

Answer 8

The first one is way off. And the second one is SOOOOOO close! I will give you the second one: it is 73,575. Tell me what you did to get your answer for a.

Answer 9

i mulitplied the numbers and that's just what i got

Answer 10

Do you know what the 2 above the 60 is? If not that is okay, I can tell you =)

Answer 11

2x

Answer 12

times 2

Answer 13

Not quite, that sign means 60 times 60. Knowing this would would be the answer?

Answer 14

3600

Answer 15

Let me break this one down for you =) \[KE=0.5 \times 75 \times 60^{2}\]\[KE=0.5*75*3600\]\[KE=75 \times 1800\]\[KE=135,000\] So B would be 135,000

Answer 16

Opps, sorry the KE should be PE.

Answer 17

if i were to c's answer what would i do with the 2 numbers. Add Subtract Multiply or Divide?

Answer 18

Now Total Mechanical Energy =PE +PEg So: \[ME=135,000 +73,575\]

Answer 19

208,575?

Answer 20

Perfect!

Answer 21

thank you

Answer 22

No problem! And just for the future, this deals with Physics so if you could post in Physics, that would be wonderful! Here is the link so you know: http://openstudy.com/study#/groups/Physics

Answer 23

okay. i'll remember that