You are standing on a bridge over a creek, holding a stone 20 feet above the water.

1. You release the stone. How long will it take the stone to hit the water?

2. You take another stone and toss it straight up with an initial velocity of 30 feet per second. How long will it take the stone to hit the water?

3. If you throw a stone straight up into the air with an initial velocity of 50 feet per second, could the stone reach a height of 60 feet above the water?

d = Vi*t + (1/2)a*t^2

In this case, acceleration is equal to -32 ft/s/s, and the displacement is 20 ft. Vi is initial velocity.

1. To find out how long it will take for the stone to hit the water after releasing it, we can use the equation of motion for objects in free fall. The equation is given by:

h = (1/2) * g * t^2

Where:
h is the height (in this case, the initial height of the stone above the water, which is 20 feet)
g is the acceleration due to gravity (approximately 32.2 feet per second squared)
t is the time it takes for the stone to hit the water

By rearranging the equation and solving for t, we get:

t = sqrt((2h) / g)

t = sqrt((2 * 20) / 32.2)

Plugging in the values, we get:

t ≈ sqrt(1.24)

t ≈ 1.11 seconds

Therefore, it will take approximately 1.11 seconds for the stone to hit the water after releasing it.

2. Similarly, to find out how long it will take for the stone to hit the water when you toss it straight up with an initial velocity of 30 feet per second, we can again use the equation of motion for objects in free fall. In this case, the initial velocity is upwards, so it will decelerate until it reaches its maximum height and then fall back down.

The equation for the time of flight of a projectile is given by:

t = (2 * v_y) / g

Where:
v_y is the vertical component of the initial velocity (30 feet per second)
g is the acceleration due to gravity (approximately 32.2 feet per second squared)

Substituting the values, we get:

t = (2 * 30) / 32.2

t ≈ 1.86 seconds

Therefore, it will take approximately 1.86 seconds for the stone to hit the water when tossed straight up with an initial velocity of 30 feet per second.

3. To determine if a stone thrown straight up into the air with an initial velocity of 50 feet per second can reach a height of 60 feet above the water, we need to calculate its maximum height using the equation:

h = (v^2) / (2 * g)

Where:
h is the maximum height
v is the initial velocity (50 feet per second)
g is the acceleration due to gravity (approximately 32.2 feet per second squared)

Substituting in the values, we get:

h = (50^2) / (2 * 32.2)

h ≈ 39.07 feet

Since the maximum height is less than 60 feet, the stone cannot reach a height of 60 feet above the water when thrown straight up with an initial velocity of 50 feet per second.

To answer these questions, we can use the basic principles of motion and the equations of motion. The key equation we will use is:

\[h(t) = h_0 + v_0t - \frac{1}{2}gt^2\]

where:
- \(h(t)\) represents the height of the stone at time \(t\)
- \(h_0\) is the initial height of the stone
- \(v_0\) is the initial velocity of the stone (positive for upward motion, negative for downward motion)
- \(g\) is the acceleration due to gravity (approximately \(32.2\) ft/s²)

1. When you release the stone, it will accelerate downward towards the water. Since the initial velocity is zero, the equation simplifies to:

\[h(t) = h_0 - \frac{1}{2}gt^2\]

Plugging in the values, where the initial height \(h_0\) is \(20\) feet and \(g\) is \(32.2\) ft/s², we can solve for \(t\) when \(h(t) = 0\) to find the time it takes for the stone to hit the water.

2. When you toss the stone straight up with an initial velocity of \(30\) ft/s, the equation becomes:

\[h(t) = h_0 + v_0t - \frac{1}{2}gt^2\]

Here, the initial height \(h_0\) is \(0\) because the stone starts at ground level. The acceleration due to gravity \(g\) is still \(32.2\) ft/s². We need to find the time it takes for the stone to reach \(h(t) = 0\) (when it hits the water). To solve for \(t\), we set \(h(t) = 0\) and solve the equation.

3. If you throw a stone straight up into the air with an initial velocity of \(50\) ft/s, we need to determine whether it can reach a height of \(60\) feet. We can use the equation:

\[h(t) = h_0 + v_0t - \frac{1}{2}gt^2\]

Here, the initial height \(h_0\) is \(0\) (ground level) and the acceleration due to gravity \(g\) is still \(32.2\) ft/s². We need to find \(t\) when \(h(t) = 60\) feet. If there exists a \(t\) that satisfies \(h(t) = 60\), it means the stone can reach a height of \(60\) feet. Otherwise, it cannot reach that height.

By solving these equations using the given values and calculating the time \(t\) for each case, we can determine the answers to the questions.