Check out some similar questions!

Meagan101 · Oct 6, 2009, 04:35 PM

If an arrow is shot upward on the moon with a velocity of 50m/s, its height in metres after t secinds is given by: h(t)= 50(t)-0.83(t)(squared). After how many seconds will the arrow hit the moon's surface? The answer is supposed to be 60.24 seconds but I get every answer BUT that. HELP PLEASE

elscarta · Oct 6, 2009, 10:22 PM

When the arrow hits the moon's surface its height = 0 so you need to solve the equation
0 = 50t - 0.83t^2
factorising a common factor of t gives
0 = t(50 - 0.83t)

this has two solutions

t= 0 or 50 - 0.83t = 0

the first solution of t = 0 matches when the arrow was initially fired upwards

the second solution when 50 - 0.83t = 0 matches when it hits the surface again.

So you need to solve
50 - 0.83t = 0
this means that
50 = 0.83t
which leads to
t = 50 / 0.83

which gives t = 60.24 s as the answer.

Hope this helps

Unknown008 · Oct 6, 2009, 11:20 PM

If you want to solve this using rate of change, OK.

$0.83t^2-50t = 0$

Differentiate:

$(2)(0.83)t - 50 = 0$

$1.66t - 50 = 0$

Solving for t is where the maximum of the equation $0.83t^2-50t = 0$ is maximum, that is the point where the arrow reaches its maximum height.

Now, solve for t:

$1.66t \cancel{- 50+50} = 0+50$

$\frac{\cancel{1.66}t}{\cancel{1.66}} = \frac{50}{1.66}$

$t = 30.12$

That is the time for half the total journey of the arrow (upwards only). Mutiply by two to get the total time of the journey :)