A toy rocket is launched vertically and remains in the air for 5.1 seconds. a) How high did the rocket get? SHOW ALL WORK. b) What was the initial launch speed? SHOW ALL WORK.
the rocket will take time to go the highest vertical height is=5.1/2sec
2.55 sec
thanks but can you explain it please? I don't really understand how to get that answer. :/
u will use the equation v=u-gt 0=u-9.8*2.55 u=9.8*2.55 then u will use the equation v^2=u^2-2* g* h
an object take equal time to go the highest position as go down to the ground
for example if a body go to the highest height by 10 sec. then it will take 10 sec for going to the ground
time period will b T=20 sec
I'm confused so I'm just going to think about this question for a while.
did u understand the time needed to go the heighest position by 2.55 sec
Well I understand that part.
time period T= 5.1 sec given
ok
I don't understand what formula should be used.
now let u= initial velocity t= 2.55 sec is needed to go to the heighest height v= final velocity=0= velocity at the heighest height
u know v=u-gt 0=u-9.8*2.55
tell me the value of u
Oh okay. I can picture the diagram in my head now. ^-^ Um give me a minute to write it out.
ok
Uh is it about 25?
ya
now r u able to calculate maximum height
I think so. Thanks. (:
which equation u wanna use to calculate highest height
Was the answer to part b) 25? or was that for part a?
b
now try to solve for part a)
Then I guess I don't know how to solve part a. Sorry.
u can use equation v^2=u^2-2* g*H 0^2=25^2-2*9.8*H
now tell me the value of H
Is it about 32?
ya
Thanks so much!
but i m not sure that u understood this math
I understand it more than before... :P
good
anyway may i know ur country name
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