Well, it seems like the ball is really falling for it! Let's break it down:
(a) If the ball is thrown horizontally off the edge of the cliff, the only force acting on it is gravity. So, we can use kinematic equations to find the time it takes for the ball to fall 20.3 m below.
Using the equation d = v_i*t + (1/2)*a*t^2, where d is the distance, v_i is the initial velocity, a is the acceleration (in this case, due to gravity, -9.8 m/s^2), and t is the time, we have:
20.3 m = 0*t + (1/2)*(-9.8 m/s^2)*t^2
Simplifying the equation, we get -4.9*t^2 = -20.3 m.
Now, let's solve for t:
t^2 = 20.3 m / 4.9 m/s^2
t^2 ≈ 4.142
Taking the square root of both sides, we get:
t ≈ √4.142
t ≈ 2.035
So, it takes approximately 2.035 seconds for the ball to fall to the ground 20.3 m below.
(b) Now, let's consider if the ball is dropped from rest off the cliff edge. In this case, the initial velocity is 0 m/s. Using the same equation as before, we have:
20.3 m = 0*t + (1/2)*(-9.8 m/s^2)*t^2
Simplifying the equation, we get -4.9*t^2 = -20.3 m.
Solving for t:
t^2 = 20.3 m / 4.9 m/s^2
t^2 ≈ 4.142
Taking the square root of both sides, we get:
t ≈ √4.142
t ≈ 2.035
So, it takes approximately 2.035 seconds for the ball to reach the ground when it is dropped from rest off the cliff edge. It's falling faster because it knows what it wants!
(c) Finally, let's imagine the ball is thrown at an initial velocity of 20.4 m/s but 12° below the horizontal. In this case, we need to consider both the vertical and horizontal components of the motion.
The initial vertical velocity can be found using sine:
v_y = v_i * sin(angle)
v_y = 20.4 m/s * sin(12°)
v_y ≈ 20.4 m/s * 0.208
v_y ≈ 4.243 m/s
Now, let's find the time it takes for the ball to fall using the vertical component of motion:
20.3 m = 0*t + (1/2)*(-9.8 m/s^2)*t^2
Simplifying the equation, we get:
-4.9*t^2 = -20.3 m
Solving for t:
t^2 = 20.3 m / 4.9 m/s^2
t^2 ≈ 4.142
Taking the square root of both sides, we get:
t ≈ √4.142
t ≈ 2.035
So, it would still take approximately 2.035 seconds for the ball to fall to the ground, even if it is thrown at an initial velocity of 20.4 m/s but 12° below the horizontal. But hey, the ball is a free-spirited traveler, exploring both the horizontal and vertical dimensions!