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!
To solve parts (a) and (b) of the problem, we need to use the kinematic equations of motion. Let's go step by step:
(a) How much later does it 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:
s = ut + (1/2)at^2
Where:
s = displacement (in this case, the height of the cliff = 95.0 m)
u = initial velocity (11.5 m/s)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
t = time
Rearranging the equation, we have:
s - (1/2)at^2 = ut
Substituting the given values:
95 - (1/2)(-9.8)(t^2) = 11.5t
Simplifying the equation, we get:
4.9t^2 + 11.5t - 95 = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
By substituting the values a = 4.9, b = 11.5, and c = -95 into the quadratic formula, you can solve for 't'. The positive value of 't' will give you the time it takes for the stone to reach the bottom of the cliff.
(b) What is its speed just before hitting?
To find the speed of the stone just before hitting the ground, we can use the equation:
v = u + at
Where:
v = final velocity
u = initial velocity (11.5 m/s)
a = acceleration (due to gravity, approximately -9.8 m/s^2)
t = time (calculated in part (a))
Substituting the values, you can calculate the final velocity 'v', which will be the speed just before hitting the ground.
Remember to utilize the correct units and round your answers appropriately.
I hope this helps you solve parts (a) and (b) of the problem! Let me know if you need any further assistance.