A man tosses a penny up into the air above a 100 ft well with a velocity of 5 ft/sec. The penny leaves his hand at a height of 4 ft. How long will it take for the penny to reach the bottom of the well?
height of penny = -16t^2 + 5t + 4
to reach bottom of well, h = -100
-16t^2 + 5t + 4 = -100
16^2 - 5t -104 = 0
using the formula:
t = appr 2.7 or a negative, which we'll reject
To answer this question, we can use the basic principles of kinematics. The motion of the penny can be divided into two parts: upward motion and downward motion.
First, let's consider the upward motion. We know the initial velocity, which is 5 ft/sec, and the displacement, which is 4 ft. Using the kinematic equation, we can determine the time it takes for the penny to reach its maximum height. The equation is:
v_f = v_i + at
where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity is 0 since the penny reaches its maximum height and momentarily stops before falling back down. The initial velocity is 5 ft/sec, and the acceleration is due to gravity, which is approximately 32 ft/sec² in this context. So the equation becomes:
0 = 5 - 32t
Solving for t, we get:
32t = 5
t = 5/32 ≈ 0.15625 seconds
Now, the second part is the downward motion. Since the penny is now in free fall, we can use the equation for the displacement of an object in free fall:
s = ut + (1/2)at²
where s is the displacement, u is the initial velocity, t is the time, and a is the acceleration. In this case, the initial velocity is 0 (since the penny is dropped from rest at its maximum height), the displacement is the total distance traveled (100 ft), and the acceleration is -32 ft/sec² (negative since it's downward).
Plugging in the values, we have:
100 = 0 + (1/2)(-32)t²
Simplifying, we get:
100 = -16t²
Dividing by -16, we obtain:
t² = -6.25
Since time cannot be negative, we realize that the equation does not have a valid real solution. This means that the penny will not reach the bottom of the well.