A rock is tossed straight up with an initial speed of 20 m/s. When it returns, it fall into a hole that is 8.0 m deep. What is the velocity of the rock (in m/s) when it hits the bottom of the hole? Use -9.80 m/s2 for the local acceleration due to gravity and assume no significant air resistance. Assume that the y axis is positive in the up direction
-8 = 0 + 20 t - 4.9 t^2
4.9 t^2 -20 t -8 = 0
solve quadratic for t
v = 20 - 9.8 t
I am sorry but I am still confused...
more details:
h = original height + Vi t - (1/2)(9.8) t^2
-8 = 0 + 20 t - 4.9 t^2
4.9 t^2 -20 t -8 = 0
solve quadratic for t
velocity = original velocity - 9.8 t
v = 20 - 9.8 t
To find the velocity of the rock when it hits the bottom of the hole, we can use the concept of free fall motion.
First, let's analyze the motion of the rock while it is in the air. When the rock is tossed straight up, it will travel upward until it reaches its highest point, and then it will fall back down.
The initial speed of the rock is 20 m/s, which means its initial velocity (v₀) is also 20 m/s. Since the rock is tossed straight up, its initial velocity is positive.
At the highest point, the rock's velocity becomes zero (v = 0 m/s) as it changes direction and starts falling downward. This point represents the peak of the motion.
Now, let's calculate the time it takes for the rock to reach its highest point. We can use the second equation of motion:
v = v₀ + at
Since the velocity becomes zero at the highest point, we can substitute v = 0 m/s:
0 = 20 m/s + (-9.80 m/s²) * t
Solving this equation for t:
-20 m/s = (-9.80 m/s²) * t
t = -20 m/s / (-9.80 m/s²) ≈ 2.04 s
So, it takes approximately 2.04 seconds for the rock to reach its highest point.
Next, let's determine the time it takes for the rock to fall from the highest point to the bottom of the hole.
Since the hole is 8.0 m deep, we need to find the distance covered by the rock during the downward fall.
We can use the first equation of motion:
s = v₀t + (1/2)at²
The initial velocity (v₀) when the rock starts falling from the highest point is 0 m/s (since its velocity becomes zero at the top).
The distance covered (s) is equal to 8.0 m (depth of the hole).
The acceleration (a) due to gravity is -9.80 m/s² (negative because it acts in the opposite direction to the upward convention).
We need to find the time (t) it takes for the rock to fall through the hole.
Rearranging the equation and plugging in the given values:
8.0 m = 0 m/s * t + (1/2)(-9.80 m/s²)t²
8.0 m = (-4.90 m/s²)t²
Dividing both sides by -4.90 m/s²:
-1.63265 s² ≈ t²
Since time cannot be negative, we take the positive square root:
t ≈ √(-1.63265) ≈ 1.28 s
Therefore, it takes approximately 1.28 seconds for the rock to fall from the highest point to the bottom of the hole.
Now, we can find the velocity (v) of the rock when it hits the bottom of the hole.
We can use the equation of motion:
v = v₀ + at
Plugging in the values:
v = 0 m/s + (-9.80 m/s²) * 1.28 s
v ≈ -12.54 m/s
Since velocity has a negative sign, indicating downward motion, we take the magnitude to get the final answer:
The velocity of the rock when it hits the bottom of the hole is approximately 12.54 m/s downward.