An object is thrown upwards and after 2 seconds is 78 m above where it was initially thrown. Assume no air resistance, and determine the initial speed the object was thrown at.
you can use this equation of motion to solve this:\[y=ut+{1\over2}at^2\]where: y = distance travelled u = initial velocity a = acceleration (which, in this case, should be equal to -g) t = time
we are looking for u right?
yes
-78m=u(2) + 1/2 (-9.8)(2)^2 right?
srry not -78
78=...
If you are learning about projectile motion, then you might find this helpful: http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html
i got some weird answer. i got -1372.8
and thank you soo much for helping
try and work out each part step-by-step. so, first what is the value of:\[t^2\]when t=2?
y = 78m t = 2 a = -9.8
yes, but your final answer is wrong, so I'm trying to take you through it one step at a time to se where you may have made a mistake.
i think i am making a mistake in the re-arranging the equation
ok, so first try rearranging the original equation:\[y=ut+{1\over2}at^2\]to get an expression for u. i.e. BEFORE plugging in the values.
y/u= t + 1/2 at^2
no - that is wrong
then isolate the u? soooo sorry im very terrible at this
think of it as:\[A=B+C\]how would you rearrange this to get an expression for B?
you need to subtract C from both sides
a-c
yes, therefore:\[y=ut+{1\over2}at^2\]first becomes:\[y-{1\over2}at^2=ut\]
next you need to divide both sides by 't'
OKAYYY!! i was trying to get the easier way out by dividing only. THANK YOU
np - do you still want me to check your final answer?
yup please. 1 min
48.8 m/s? :s
yes - you've got it right - well done!
LOL THANK YOU SOOO MUCH!!! YOU EXPLAIN VERY GOOD
yw - I'm glad I was able to help :)
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