Question

Please help !!
A student leans out of her fourth-floor window and throws an ice cube straight up with an initial velocity of 24 feet per second. Her window is 50 feet above the pavement.

Answer 1

What is the function given the height of the ice cube (t) seconds after she throws it
?

Answer 2

could you clarify what they are asking for i.e. distance, velocity, what units etc..

Answer 3

you should use the kinematics equations. \[\ y_{final} = y _{initial}+v _{y}t -\frac{ 1 }{ 2 } g t ^{2}\]. Then you get \[y _{final} = 50 + 24t-16t ^{2}\].
Just replace each of the constants in the equation with what you are given.

Answer 4

Where did you get the 16 ?

Answer 5

\[g=9.81m/s^2=32.2ft/s^2\]

Answer 6

But the word problem did not have that number in it ?

Answer 7

How do I calculate how fast it is hitting the ground ?

Answer 8

That's the acceleration of the object due to gravity - the problem might have assumed that you had already been given that information ^_^

To calculate how fast it hits the ground, you can use the 4th equation of motion

\[v_f^2 = v_i^2+2a\Delta y\]

where delta y is the displacement from the original position - in this case, the you start at y=50ft and end at y=0ft, so the delta y is a negative number; g is also a negative acceleration.

Answer 9

is this what it will look like 0=50+2adelta y

Answer 10

the equation
\[ v_f^2 = v_i^2 + 2a\Delta y\]
Is to find the final velocity as it hits the ground - you already know the delta y; it starts at y=50ft and ends at y=0ft

So it would be
\[v_f^2 = (24 \ ft/s)^2 + 2 (-32 \ ft/s^2)(0 ft - 50 ft)\]
\[ v_f^2 = (24\ ft/s)^2 + 2(32 \ ft/s^2)(50ft)\]
and remember, the final velocity will be going in the negative y direction.

The formula to find distance as a function of time is what @heyaa posted above
\[y(t) = (24 \ ft/s)t - \frac{1}{2}(32 \ ft/s^2)t^2\]