Question

a rock thrown downward with an unknown intial velocity from a height of 180 ft reaches the ground in 5s. find the velocity of the rock when it hits the ground. (acceleration due to gravity

Answer 1

The velocity of the rock when it hits the ground is -49 m/s (downwards).

Step-by-step explanation:

To find the velocity of the rock when it hits the ground, we can use the equation:

V = u + at

Where:

• V is the final velocity (which is what we're trying to find)
• u is the initial velocity (unknown in this case)
• a is the acceleration due to gravity (-9.8 m/s^2)
• t is the time taken for the rock to reach the ground (5s)

Assuming the rock was thrown downwards, the initial velocity can be considered as negative.

Plugging in the values:

V = -9.8 m/s^2 × 5s = -49 m/s

Therefore, the velocity of the rock when it hits the ground is -49 m/s (downwards).

Answer 2

The velocity of the rock when it hits the ground is approximately -161 ft/s. The negative sign indicates that the velocity is in the downward direction.

To find the velocity of the rock when it hits the ground, you can use the following kinematic equation for objects in free fall under the influence of gravity:

v = u + at

Where:

- v is the final velocity (velocity when the rock hits the ground).

- u is the initial velocity (which is unknown in this case).

- a is the acceleration due to gravity (approximately -32.2 ft/s², considering downward motion as negative).

- t is the time (given as 5 seconds).

You are given that the rock was thrown downward from a height of 180 ft, which means it starts from rest at that height. Since it's starting from rest, the initial velocity u is 0 ft/s.

Now, plug the values into the equation:

`\[v = 0 + (-32.2 \, \text{ft/s}²) \cdot (5 \, \text{s})\]`

`\[v = -161 \, \text{ft/s}\]`