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

1)
y = yo + voyt + 0.5at^2

0 = 75 + 12t - 9.80t^2 (after this, use quadratic formula)

t = 3.4 seconds

2)
v = vo + at

v = 12 + (9.80)(3.4)

v = 45 m/s

3)
v = vo + at

0 = 12 -9.80t

9.80t = 12

t = 1.2 seconds to get up to it's max. height.

y = yo + voyt + 0.5at^2

y = 75 + 12(1.2) - 0.5(1.2)^2

y = 75 + 14 - 0.7

y = 88 m

dY = 88-75 = 13 m

13 * 2 = 26 m

Total distance traveled = 75 + 26 = 81 m.

Answer 2

You need two equations. The formula is h(t) = 1/2 a t^2 + v t + h0

You are given v=12 m/s and h0=75m. The acceleration (a) is from gravity and is -9.8 m/s^2. It is negative because gravity is pulling down. T is time and h(t) is the height at that time.

The second equation is v(t) = v0 + at where v(t) is the velocity at time t.

1) Solve the first equation for h = 0
2) Use the time in (1) in the second equation to find the velocity at that time.
3) Use the second equation to find the time when the velocity is zero. That is when the stone stopped going up and started to come down. Then use that time in the first equation to find what height that occurred at. Since the stone starts at 75 m, rises to that height, then falls to the ground, the distance traveled is twice that height - 75.

Good luck.

Answer 3

The height of the stone is given by

s(t) = -1/2 gt^2 + v0*t + s0

v0=initial speed, 12 m/s
s0=initial height, 75 m
g=gravity, 9.8 m/s^2

1) plug in those numbers and solve s(t)=0, using the quadratic formula. You'll get two values of t, take the positive one

2) Plug that value of t into the velocity at time t, given by v(t) = -gt + v0

3) total distance: need the total distance up, and down. The maximum height occurs when v(t)=0 (at first, it's positive, for going up, then it switches to negative for going down). Solve -gt+v0=0 to find the time at which it stops going up.

Then, plug that value of t into s(t) to find the actual maximum height. That value is the distance that it fell, and you can get the distance going up by subtracting the initial height of 75.