t = (-u +/- (sqrt) u^2+2as)/a?

Question

I have the above formula. u = 10m/s , a=-9.81 s = (distance)

Q1. Need to find time (t in seconds) that it takes for an object to be thrown up in the air 4 metres.

I think answer to Q1 is either 0.546 or 1.4922 can ne one confirm?

Q2. Says calculate time it takes to fall back down to earth. Havent a clue how to do this part can ne one help me? I tried just making "a" positive so that it equaled gravity of 9.81 but this doesnt seem to work

Need answer by Monday please help me
Thanks

Answer 1

The 4 kinematic formulae are:
v=u+at
v^2=u^2+2as
s=ut+(1/2)at^2
s=(1/2)(u+v)(t)

Q1. t=?, u=10m/s, a=-9.81s, s=4m
Use s = ut + (1/2)at^2.
4 = 10t + (1/2)(-9.81)t^2
4.905t^2 - 10t + 4 = 0
t = [10 +- sqrt(10^2 - 4(4.905)(4)] / [(2)(4.905)]
t = [10 +- sqrt(21.52)] / [9.81]
t = 1.49s, 0.546s (to 3 significant figures)
Yes, your answer's correct. :D

Q2. When thrown up, t=?, u=10m/s, a=-9.81s, v=0m/s
(At the highest point, the velocity is zero as it is about to come back down.)
v = u + at
0 = 10 - 9.81t
t = 10/9.81
t = 1.019s
Hence the time taken for the object to go up and fall back down to earth is 1.019 x 2 = 2.038s as the time taken for the upward and downward journeys are the same.

If they only require the time for the object to fall back down and not the time taken for it to go up, it will 1.019s.

Prove that time for going up is the same as time for coming down.
s = (1/2)(u+v)t
s = (1/2)(10+0)(1.019)
s = 5.095m
(This is the highest that the object will go with the velocity of 10m/s.)

When coming down, t=?, u=0m/s, a=9.81s, s=5.095m
s = ut + (1/2)at^2
5.095 = 0 + (1/2)(9.81)(t^2)
t^2 = 5.095/(1/2)(9.81)
t = 1.019s (It is the same as that for going up to its highest point.)

Answer 2

I'm not sure whether you were given this equation or if you already manipulated it yourself, but i think it would be helpful to point out that it comes from x=(1/2)a*t^2+v0t+x0

x is the vertical position of the object
a is the constant acceleration caused by earth's gravity
t is the time
v0 is the initial velocity
x0 is the initial vertical position

in your problem the initial vertical position is 0, so the proper equation is
x=(1/2)a*t^2+v0t
if you solve for t via the quadratic formula you obtain the equation that you have intially. that was the hard part which obviously you had already done or someone had done for you. now simply set x=0 (or in the case of your equation, s=0) in your and solve for t.

The answer for Q2 is:
t=sqrt(10^2+2(-9.8)*0)-10/-9.8 = (10-10)/-9.8 = 0
and
t= -sqrt(10^2+2(-9.8)*0)-10/-9.8 = (-10-10)/ -9.8 = 2.04s

and that's your answer. there's plenty of good information out there about the kinematics of constant acceleration, if you just learn what is going on physically it makes it much less abstract.

by the way, you are exactly right on Q1, it's both. The object ascends past 4 meters, hits a peak (of 5.1 meters at 1.02 seconds, but i digress) and then decends past 4 meteres again. thus there are two values for t for any height between 0 and 5.1 meters.

Answer 3

Q1. The answer is 0.903s
Q2. It takes another 0.903s to fall back to the ground.

Answer 4

The anser is 21

Answer 5

E=Mc^2 Divided by the meaning of life which is 42.
That is the answer to everything ... use it as you like with my permission :-)

Answer 6

The answer is def 37.5

Answer 7

...aint that cheating...good on ya...hope you get an a* grade...happy hols...

Answer 8

Sorry, can't help

Answer 9

absolutely no idea - but thanks for the 2 points!

it went straight over my head at t = !!!!!!