A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 16 m/s. The cliff is h = 60 m above the water's surface, as shown below. How long does it take for the stone to fall to the water? With what speed does it strike the water?
h = Vo*t + 0.5g*t^2 = 60 m.
16t + 4.9t^2 = 60
4.9t^2 + 16t - 60 = 0
Use Quad. Formula.
Tf = 2.23 s. = Fall time.
V = Vo + gt = 16 + 9.8*2.23 = 37.86 m/s.
To find the time it takes for the stone to fall to the water, we can use the kinematic equation for vertical motion:
h = vi * t + (1/2) * g * t^2
Where:
- h is the height of the cliff (60 m)
- vi is the initial vertical velocity of the stone (16 m/s)
- t is the time taken for the stone to fall
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
First, rearrange the equation to solve for time (t):
(1/2) * g * t^2 + vi * t - h = 0
This is a quadratic equation, so we can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = (1/2) * g, b = vi, and c = -h. Plug in the values:
t = (-vi ± √(vi^2 - 4 * (1/2) * g * (-h))) / (2 * (1/2) * g)
t = (-16 ± √(16^2 - 4 * (1/2) * 9.8 * (-60))) / (2 * (1/2) * 9.8)
You'll get two possible values for t since we used the ± sign. Make sure to choose the positive value since time cannot be negative.
Once you have the value of t, you can calculate the speed at which the stone strikes the water using the equation:
v = vi + g * t
Plug in the values:
v = 16 + 9.8 * t
Evaluate the expression to find the final answer.
It's worth noting that air resistance has not been considered in this calculation, so the actual speed of the stone hitting the water might be slightly different in real-world conditions.