Question

A student drops a rock from a bridge to the

water 12.2 m below.
With what speed does the rock strike
the water? The acceleration of gravity is
9.8 m/s
2
.
Answer in units of m/s.

Answer 1

The rock strikes the water at a speed of approximately 15.46 m/s, calculated using the kinematic equation for uniformly accelerated motion with the acceleration due to gravity and the distance fallen.

Step-by-step explanation:

The problem asks to determine the speed at which the rock strikes the water after being dropped from a height. To calculate this, we can use the kinematic equation for uniformly accelerated motion, which is:

v^2 = u^2 + 2as

Where:

• v is the final velocity,
• u is the initial velocity (which is zero in this case, since the rock is dropped),
• a is the acceleration due to gravity (9.8 m/s^2), and
• s is the distance fallen (12.2 m).

Substitute the known values into the equation:

v^2 = 0^2 + 2 * 9.8 m/s^2 * 12.2 m

v^2 = 239.12 m^2/s^2

Now we take the square root to find the final velocity:

v = √239.12 m^2/s^2

v ≈ 15.46 m/s

So, the speed at which the rock strikes the water is approximately 15.46 m/s.