Jeff is standing on a bridge 5M above the river, He picks up a small rock and tosses it into the air, the rock reaches a height of 12M in one second and hits the water in about 2.3 seconds Determine when the rock reaches a height equal to the bridge but on the way down.
hf=hi+vi*t-1/2 9.8 t^2
vf=vi-9.8t at the top, vf=0, t=1
vi=9.8m/s
then
hf=hi+vi*t-4.9t^2
but hf=hi, so
4.9t^2-vi*t=0
t(4.9t-vi)=0 or t=9.8/4.9=2seconds
which is curious, and you need to think about it. It took just as long to go up as down...
To determine when the rock reaches a height equal to the bridge on the way down, we can use the principles of projectile motion.
Let's break down the information given in the problem:
1. Jeff tosses the rock into the air, and it reaches a height of 12 meters in 1 second.
2. The rock hits the water in about 2.3 seconds.
3. The bridge is 5 meters above the river.
To solve this problem, we need to find the time it takes for the rock to reach the height of the bridge while coming down. We can use the time it takes for the rock to reach its maximum height.
First, let's determine the initial velocity of the rock when it is thrown into the air.
Using the formula:
v = u + at
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity
a = acceleration due to gravity (-9.8 m/s^2)
t = time (1 second)
Rearranging the formula gives us:
u = v - at
Plugging in the values, we get:
u = 0 - (-9.8 * 1)
u = 9.8 m/s
Now, since we know the initial velocity, we can calculate the time it takes for the rock to reach its maximum height using the formula:
v = u + at
Where:
v = final velocity (0 m/s at the highest point)
u = initial velocity (9.8 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time (unknown)
Rearranging the formula gives us:
t = (v - u) / a
Since the final velocity is 0 m/s, the equation becomes:
t = -u / a
Plugging in the values, we get:
t = -9.8 / -9.8
t = 1 second
Therefore, it takes 1 second for the rock to reach its maximum height.
Now, we need to calculate the time it takes for the rock to fall from its maximum height to the height of the bridge. Since the heights are given in meters, we can use the equation:
s = ut + (1/2)at^2
Where:
s = distance (height of the bridge = 5 meters)
u = initial velocity (9.8 m/s)
t = time (unknown)
a = acceleration due to gravity (-9.8 m/s^2)
Substituting the known values and solving for time, we have:
5 = 9.8t + (1/2)(-9.8)(t^2)
Rearranging and simplifying:
-4.9t^2 + 9.8t + 5 = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -4.9, b = 9.8, and c = 5. Plugging in the values, we get:
t = (-9.8 ± √(9.8^2 - 4(-4.9)(5))) / (2 * -4.9)
Simplifying further, we find:
t ≈ 1.52 seconds or t ≈ 0.681 seconds
Since we are interested in the time it takes for the rock to reach a height equal to the bridge, we can disregard the negative value solution. Thus, the rock reaches a height equal to the bridge, while on the way down, in approximately 0.681 seconds.