A startled armadillo leaps upward, rising 0.515 m in the first 0.190 s. (a) What is its initial speed as it leaves the ground? (b) What is its speed at the height of 0.515 m? (c) How much higher does it go? Use g=9.81 m/s2.
To answer these questions, we can use the equations of motion under constant acceleration. Let's go through each part one by one.
(a) To find the initial speed of the armadillo, we need to use the equation:
v = u + at
Where:
v is the final velocity (which is zero when the armadillo reaches its peak height)
u is the initial velocity (what we're trying to find)
a is the acceleration (which is the acceleration due to gravity, -9.81 m/s^2 since it's acting against the upward motion)
t is the time taken to reach the peak height (given as 0.190 s)
Substituting the known values into the equation, we have:
0 = u - 9.81 * 0.190
Rearranging the equation to solve for u, we get:
u = 9.81 * 0.190
Calculating this, we find that the initial speed of the armadillo is approximately 1.8619 m/s.
(b) To determine the speed of the armadillo at a height of 0.515 m, we need to use the equation:
v^2 = u^2 + 2aΔy
Where:
v is the final velocity (what we're trying to find)
u is the initial velocity (from part (a), which is 1.8619 m/s)
a is the acceleration (which is -9.81 m/s^2)
Δy is the change in height (0.515 m)
Substituting the known values, we have:
v^2 = (1.8619)^2 + 2 * (-9.81) * 0.515
Simplifying the equation, we get:
v^2 = 3.4676 - 10.0689
v^2 = -6.6013
Since we cannot have a negative speed, this means that the armadillo does not reach a height of 0.515 m.
(c) Finally, we need to determine how much higher the armadillo actually goes. To do this, we can start by finding the time it takes for the armadillo to reach its peak height.
Using the equation: Δy = ut + (1/2)at^2
Where:
Δy is the change in height (0.515 m)
u is the initial velocity (from part (a), which is 1.8619 m/s)
a is the acceleration (which is -9.81 m/s^2)
t is the time taken to reach the peak height (what we're trying to find)
Substituting the known values, we have:
0.515 = 1.8619 * t + (1/2) * (-9.81) * t^2
Simplifying the equation, we get:
-4.905t^2 + 1.8619t + 0.515 = 0
Using the quadratic formula, we can solve for t. The quadratic formula is given by:
t = (-b ± √(b^2 - 4ac)) / (2a)
Applying this formula to our quadratic equation, we find:
t ≈ 0.246 s (positive value)
Therefore, the armadillo takes approximately 0.246 seconds to reach its peak height.
To calculate how much higher it goes, we can substitute this value of t back into the equation for height:
Δy = ut + (1/2)at^2
Δy ≈ 1.8619 * 0.246 + (1/2) * (-9.81) * (0.246)^2
Therefore, the armadillo goes approximately 0.5662 meters higher than the initial height.
In summary:
(a) The initial speed of the armadillo is approximately 1.8619 m/s.
(b) The armadillo does not reach a height of 0.515 m.
(c) The armadillo goes approximately 0.5662 m higher than its initial height.