A stone is thrown vertically upward with a speed of 11.5 m/s from the edge of a cliff 95.0 m high.

(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?
For (c), the right answer is 108.46 m. I really need help on parts (a) and (b) please!

Sure! Let's break down the problem into parts and find the solutions step by step:

(a) How much later does the stone reach the bottom of the cliff?

To find the time it takes for the stone to reach the bottom of the cliff, we need to use the equation of motion for vertical motion:

s = ut + (1/2)at^2

Where:
s = displacement (final position - initial position)
u = initial velocity
t = time taken
a = acceleration due to gravity

In this case, the stone is thrown upward, so the initial velocity (u) is 11.5 m/s and the acceleration due to gravity (a) is -9.8 m/s^2 (negative because it acts downward). The displacement (s) is the negative height of the cliff, -95.0 m.

Therefore, the equation becomes:

-95.0 = (11.5)t + (1/2)(-9.8)t^2

Now, we can solve this quadratic equation for t.

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)

Where a = (1/2)(-9.8), b = 11.5, and c = -95.0.

By substituting these values into the formula, we get two possible solutions for t. We discard the negative solution since time cannot be negative in this context:

t = (-11.5 ± √(11.5^2 - 4(0.5)(-95.0)))/(2(0.5))

Simplifying the equation further:

t = (-11.5 ± √(132.25 + 190))/(1)
t = (-11.5 ± √322.25)/(1)
t = (-11.5 ± 17.95)/(1)

Now we calculate both possible values of t:

t1 = (-11.5 + 17.95)/(1) = 6.45 s
t2 = (-11.5 - 17.95)/(1) = -29.45 s (discarded since it's negative)

Therefore, the stone takes approximately 6.45 seconds to reach the bottom of the cliff.

(b) What is the speed of the stone just before hitting?

To find the speed just before hitting the ground, we can use the equation:

v = u + at

Where:
v = final velocity (speed just before hitting)
u = initial velocity (11.5 m/s, upward)
a = acceleration due to gravity (-9.8 m/s^2, downward)

Plugging in the values:

v = 11.5 + (-9.8)(6.45)
v = 11.5 - 63.06
v = -51.56 m/s

The negative sign indicates that the velocity is directed downward. To get the magnitude of the speed, we take the absolute value:

Speed = |-51.56 m/s| = 51.56 m/s

Therefore, the speed of the stone just before hitting the ground is approximately 51.56 m/s.

Thus, we have solved parts (a) and (b) of the problem. If you have any further questions or need clarification, feel free to ask!