A stone is thrown vertically upward with a speed of 17m/sec from the edge of a cliff 90 m high.

a.)How much later does it reach the bottom of the cliff?

b.) what is it’s speed just before hitting?

c.) what total distance did it travel?

To solve this problem, we need to break it down into parts and apply the principles of kinematics. We have a stone thrown upward from a height of 90 meters with an initial speed of 17 m/s. We'll analyze the motion in two phases: the upward motion until it reaches its peak and the downward motion until it hits the ground. Let's tackle each part step by step.

Part A: Time to Reach the Bottom of the Cliff

First, we need to determine how long it takes for the stone to reach the bottom of the cliff. The total height from which the stone is thrown is 90 meters, and we need to find the time it takes to go up and then come down.

1. Time to Reach the Peak

When the stone is thrown upward, it will eventually stop rising and start falling back down. The time to reach the peak can be calculated using the formula:

t_peak = v_initial / g

Where:

• v_initial = 17 m/s (initial speed)
• g = 9.81 m/s² (acceleration due to gravity)

Plugging in the values:

t_peak = 17 m/s / 9.81 m/s² ≈ 1.73 seconds

2. Maximum Height Above the Cliff

Next, we calculate how high the stone goes above the cliff before it starts to fall back down. We can use the formula:

h_peak = (v_initial²) / (2g)

Substituting the values:

h_peak = (17 m/s)² / (2 * 9.81 m/s²) ≈ 14.5 meters

So, the total height from the ground is:

h_total = 90 m + 14.5 m = 104.5 m

3. Time to Fall from the Maximum Height

Now, we need to find the time it takes to fall from this maximum height back to the ground. We can use the formula for free fall:

h = (1/2)gt²

Rearranging gives:

t_fall = √(2h/g)

Substituting the total height:

t_fall = √(2 * 104.5 m / 9.81 m/s²) ≈ 4.58 seconds

4. Total Time

The total time taken to reach the bottom of the cliff is the sum of the time to reach the peak and the time to fall back down:

t_total = t_peak + t_fall ≈ 1.73 s + 4.58 s ≈ 6.31 seconds

Part B: Speed Just Before Hitting the Ground

To find the speed just before the stone hits the ground, we can use the following kinematic equation:

v_final² = v_initial² + 2gh

Here, the initial speed when falling starts is 0 m/s (at the peak), and we are considering the total height of 104.5 m:

v_final² = 0 + 2 * 9.81 m/s² * 104.5 m

Calculating gives:

v_final² = 2057.49 m²/s²

Taking the square root:

v_final ≈ 45.3 m/s

Part C: Total Distance Traveled

The total distance traveled by the stone includes both the upward distance and the downward distance. The upward distance is the height it reached above the cliff, and the downward distance is the total height from the cliff to the ground.

Distance_up = h_peak = 14.5 m

Distance_down = h_total = 104.5 m

Thus, the total distance traveled is:

Total Distance = Distance_up + Distance_down = 14.5 m + 104.5 m = 119 m

Summary of Results

• Time to reach the bottom of the cliff: 6.31 seconds
• Speed just before hitting the ground: 45.3 m/s
• Total distance traveled: 119 meters

By breaking down the problem into manageable parts, we can clearly see how the stone's motion is influenced by gravity and its initial velocity. This approach not only helps in solving the problem but also deepens our understanding of kinematic principles.