Question

Eventually, a stone thrown vertically downward from a high bridge strikes the water with a speed of 49.0 m/s. If the initial speed was 19.6m/s, how high is the bridge? How long was the stone in the air?

Answer 1

To determine the height of the bridge and the time the stone was in the air, we can use the equations of motion for vertical motion.

Let's denote the initial velocity of the stone as $u$ (19.6 m/s), the final velocity as $v$ (-49.0 m/s, negative because it's in the downward direction), the acceleration as $a$ (-9.8 m/s², acceleration due to gravity), the time taken as $t$, and the height of the bridge as $h$.

We can use the equation for final velocity:

$v=u+at$

Substituting the given values:

$-49.0=19.6-9.8t$

Simplifying the equation:

$-9.8t=-49.0-19.6$

$-9.8t=-68.6$

Dividing both sides by -9.8:

$t=\frac{-68.6}{-9.8}$

$t\approx 7.0\text{seconds}$

So, the stone was in the air for approximately 7.0 seconds.

To find the height of the bridge, we can use the equation for distance (height):

$h=ut+\frac{1}{2}$