A startled armadillo leaps upward, rising 0.644 m in 0.170 s.
a. What is its initial speed?
b. What is its speed at this height?
c. How much higher does it go?
For a. 0 is not the answer and neither is 3.79 m/s (product of .644m*.170s)
a. h = Vo*t + 0.5g*t^2 = 0.644 m.
Vo*0.17 - 4.9*0.17^2 = 0.644.
0.17Vo = 0.644 + 0.142 = 0.786.
Vo = 4.62 m/s.
b. V=Vo + g*t=4.62 - 9.8*0.17 = 2.95 m/s.
c. Vo^2 = Vo^2 + 2g*h = 0.
h = -Vo^2/2g = -(4.62^2)/-19.6 = 1.09 m.
1.09-0.644 = 0.446 m higher.
To solve these questions, we will use the equations of motion and the principles of kinematics.
a. To find the initial speed, we can use the equation:
v = u + at
Where:
v = final velocity (unknown in this case)
u = initial velocity (unknown in this case)
a = acceleration (which is equal to -9.8 m/s^2 for objects in free fall)
t = time = 0.170 s
Since the armadillo is leaping upward, it is moving against gravity, so we take acceleration as a negative value (-9.8 m/s^2).
Rearranging the equation, we have:
u = v - at
Substituting the values we know:
u = 0 - (-9.8 m/s^2)(0.170 s)
Calculate the value:
u = 1.666 m/s (rounded to three decimal places)
Therefore, the initial speed of the armadillo is approximately 1.666 m/s.
b. To find the speed at the maximum height, we know that at the highest point, the velocity will momentarily become 0. This happens because the armadillo reaches its peak and changes direction.
Using the formula:
v = u + at
We want to find v (final velocity) when u (initial velocity) is 1.666 m/s, and a (acceleration) is -9.8 m/s^2.
Rearranging the equation:
v = 1.666 m/s - (-9.8 m/s^2)(0.170 s)
Calculate the value:
v = 3.380 m/s (rounded to three decimal places)
Therefore, the speed of the armadillo at the maximum height is approximately 3.380 m/s.
c. To find how much higher the armadillo goes, we will use the equation for displacement:
s = ut + (1/2)at^2
We want to find s (displacement), u (initial velocity), a (acceleration), and t (time).
Substituting the known values, we have:
s = (1.666 m/s)(0.170 s) + (1/2)(-9.8 m/s^2)(0.170 s)^2
Calculate the value:
s = 0.13858 m (rounded to five decimal places)
Therefore, the armadillo goes approximately 0.139 m (or 13.9 cm) higher.