A girl standing on a bridge throws a stone vertically downward with an initial velocity of 15.0 m/s into the river below. If the stone hit's the water 2.00 seconds later what is the height of the bridge above the water?
h = Vo*t + 0.5g*t^2
To find the height of the bridge above the water, we can use the equations of motion and the kinematic equation for vertical motion:
h = h0 + v0t - 1/2gt^2
Given:
Initial velocity, v0 = 15.0 m/s (negative due to downward direction)
Time, t = 2.00 s
Acceleration due to gravity, g = 9.8 m/s^2 (negative due to upward direction)
Substituting the values into the equation of motion, we have:
h = 0 + (-15.0 m/s)(2.00 s) - 1/2(9.8 m/s^2)(2.00 s)^2
Now we can calculate the height of the bridge above the water.
h = -30.0 m - 1/2(-19.6 m/s^2)(4.00 s^2)
h = -30.0 m + 39.2 m
h = 9.2 m
Therefore, the height of the bridge above the water is 9.2 meters.
To find the height of the bridge above the water, we need to determine the distance the stone traveled vertically downward during the 2.00 seconds.
We know that the initial velocity (Vi) of the stone is 15.0 m/s and the time (t) is 2.00 seconds. Since the stone is thrown straight down, the acceleration due to gravity (g) can be taken as -9.8 m/s² because the arrow notation (\(-\)) indicates downwards.
First, let's find the distance traveled by the stone using the equation:
\[d = V_i \cdot t + \frac{1}{2} \cdot g \cdot t^2\]
Substituting the given values into the equation:
\[d = 15.0 \, \text{m/s} \cdot 2.00 \, \text{s} + \frac{1}{2} \cdot (-9.8 \, \text{m/s}^2) \cdot (2.00 \, \text{s})^2\]
Solving this equation step by step:
\[d = 30.0 \, \text{m} + \frac{1}{2} \cdot (-9.8 \, \text{m/s}^2) \cdot 4.00 \, \text{s}^2\]
\[d = 30.0 \, \text{m} + (-19.6 \, \text{m})\]
\[d = 10.4 \, \text{m}\]
Therefore, the height of the bridge above the water is 10.4 meters.