a rock is dropped from the top of a tall building. the rocks displacement in the last second before it hits the ground is 45% of the entire distance it falls. how tall is the building

## Answers (3)

The equation for all acceleration questions is d=(1/2)at^2. For falling objects that becomes h=16t^2.

The equation for the first part of the fall is: 0.55h = 16(t - 1)^2

So we have two equations with two unknowns. Since the first equation is already solved for h, you can simply substitute that for h in the second equation and crank out a value for t.

0.55(16t^2) = 16t^2 - 32t + 16

0.55t^2 = t^2 - 2t + 1

0.45t^2 - 2t + 1 = 0 Use the quadratic equation now.

d=(1/2)at^2

For falling objects d=h and a=32 ft/sec^2 so 1/2 x 32 = 16

You really over complicated this.

I didn't, it is complicated right from the start.

Since it's a free fall problem, we use y for position, acceleration for gravity, g=9.8

and I will not use the negative sign for gravity for the sake of simplicity.

initial velocity& position are vi, yi =0, so omit them from the kinematics equations.

dt=change in time, yf= final position.

using y=yi+vi*dt+ (g*dt^2)/2

y=(g*dt^2)/2, keep this in mind.

Now we are going to apply the situation we got.

yi-------------------->|----------->|yf

yi-------------------y------------->|yf

when the distance/position is simply y, and time= t

while, what we want is....

yi--------y*0.55-----|yf

when (t-1).

this!

So, our equation will look like this,

0.55*y=(g*(t-1)^2)/2

Now, solve for (t-1)^2.

(t-1)^2=(2*0.55*y)/g, when y=(g*t^2)/2, as you see g and #2 will cancel out.

plug (g*t^2)/2 in to y. So we only have 1 variable, t. and solve for t

(t^2-2t+1)= 0.55t^2

.45t^2-2t+1=0

put those numbers in the quadratic program to solve for t.

I got 3.87 and 0.574. but (0.574 -1) is less than 0 which doesn't make sense because time should be greater or equal to 0.

use time 3.87s to solve the rest.

This time is the total time it took to hit the ground.

So we know g,t to solve y. awesome!

y= (g*t^2)/2

y= (9.8*3.87^2)/2

y=73.4m

good luck on your homework guys!

Equation 1:

D = (1/2)* (a) * (t^2)

D= (4.9)*(t^2)

Equation 2:

0.55D = (1/2) * a *(t-1)^2

0.55D = 4.9*(T^2 - 2T +1)

plug 1 into 2:

(0.55)*(4.9)*(t^2)= (4.9)*(T^2 - 2T +1)

cross out the common factors (4.9):

0.55T^2 = T^2 - 2T +1

Move everything to 1 side

0= T^2 - 0.55T^2 - 2T +1

Final equation:

0 = 0.45T^2 - 2T +1

quadratic formula..... T= 3.87

Plug T back into equation 1

D = 4.9 *(3.87^2)

D=73.4 m

I made an account just to answer this question lool all the answers i found before were useless

where did you get the 16 from in the first equation