please please please! i need help with this question!!?

Question

a stone is thrown vertically upward with a speed of 12 m/s from the edge of a cliff 75m high.

1) How much later does it reach the bottom of the cliff?

2) What is its speed just before hitting?

3) What total distance did it travel?

im having so much trouble with this question, u have no idea! im hoping you guys can help me !

thnx for all of your help!

Answer 1

First, how long till it stops going up?
v=u+at
0=12m/s+9.8t
hence t=12/9.8 (the minuses cancel)
hence it takes 1.225 seconds to stop going up.

Now, how far did it travel up?
s=ut+1/2at^2
s=12 x 1.225+(1/2 x 9.8 x 1.500)
hence s=22.053m

So, answer for (3) is 97.053m total distance travelled

Now we know it took 1.225sec to go up, so next we find how long it took to fall once it started coming back down
s=ut+1/2at^2
97.053=(0 x t)+(1/2 x 9.8 x t^2)
hence t^2=19.8067
then t=4.450

Therefore answer to (1) it reaches the bottom of the cliff 5.675 seconds later.

Now to find out how fast it was travelling just before it hit, use the time we just worked: t = 4.450
v=u+at
v=0+(9.8 x 4.450)
then v=43.61m/s, the answer for (2)

I hope this helps you!!!

Answer 2

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Answer 3

x=xi+v*t-.5*9.8*t^2
0=75+12*t-4.6t^2
t=5.547 seconds
v=12-5.547*9.8=-42.3 seconds
The total distance traveled is 90.64 meters