To solve this problem, we need to use the equations of motion.

First, let's find the time it takes for the dropped rock to hit the ground. We know that the height (h) is 3.0 × 102 m, the initial velocity (u) is 0 (because it's dropped), and the acceleration (a) is 9.8 m/s² (due to gravity). We can use the equation:

h = ut + 0.5at²

Substituting the known values, we get:

3.0 × 102 = 0*t + 0.5*9.8*t²

Solving for t (time), we get t = sqrt((2*3.0 × 102)/9.8) = 24.8 seconds.

Now, let's find the time it takes for the thrown rock to hit the ground. This time, the initial velocity (u) is 10.0 m/s. Using the same equation:

3.0 × 102 = 10.0*t + 0.5*9.8*t²

This is a quadratic equation, and solving for t, we get t = 19.6 seconds.

So, the thrown rock hits the ground 24.8 - 19.6 = 5.2 seconds earlier than the dropped rock.

Question

To solve this problem, we need to use the equations of motion.

First, let's find the time it takes for the dropped rock to hit the ground. We know that the height (h) is 3.0 × 102 m, the initial velocity (u) is 0 (because it's dropped), and the acceleration (a) is 9.8 m/s² (due to gravity). We can use the equation:

h = ut + 0.5at²

Substituting the known values, we get:

3.0 × 102 = 0*t + 0.5*9.8*t²

Solving for t (time), we get t = sqrt((2*3.0 × 102)/9.8) = 24.8 seconds.

Now, let's find the time it takes for the thrown rock to hit the ground. This time, the initial velocity (u) is 10.0 m/s. Using the same equation:

3.0 × 102 = 10.0*t + 0.5*9.8*t²

This is a quadratic equation, and solving for t, we get t = 19.6 seconds.

So, the thrown rock hits the ground 24.8 - 19.6 = 5.2 seconds earlier than the dropped rock.

Knowee AI · Accepted Answer

To solve this problem, we need to use the equations of motion.

First, let's find the time it takes for the dropped rock to hit the ground. We know that the height (h) is 3.0 × 102 m, the initial velocity (u) is 0 (because it's dropped), and the acceleration (a) is 9.8 m/s² (due to gravity). We can use the equation:

h = ut + 0.5at²

Substituting the known values, we get:

3.0 × 102 = 0*t + 0.5*9.8*t²

Solving for t (time), we get t = sqrt((2*3.0 × 102)/9.8) = 24.8 seconds.

Now, let's find the time it takes for the thrown rock to hit the ground. This time, the initial velocity (u) is 10.0 m/s. Using the same equation:

3.0 × 102 = 10.0*t + 0.5*9.8*t²

This is a quadratic equation, and solving for t, we get t = 19.6 seconds.

So, the thrown rock hits the ground 24.8 - 19.6 = 5.2 seconds earlier than the dropped rock.

At the same moment from the top of a building 3.0 × 102 m tall, one rock is dropped from rest and one is thrown directly downward with an initial speed of 10.0 m/s.How much EARLIER does the thrown rock strike the ground?Neglect air resistance and express your answer in seconds.

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