A startled armadillo leaps upward, rising 0.596 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 0.596 m? (c) How much higher does it go?
hf(t)=vi*t- 1/2 g t^2
.596=vi*.190- 4.9 (.190)^2
solve for vi
b) vf=Vi-gt t=.190, Vi above
c) how high?
Vf at top is zero.
Vf^2=Vi^2+2gd solve for d.
To solve this problem, we will use the equations of motion for vertical motion under constant acceleration. Let's break down the problem step by step.
(a) What is its initial speed as it leaves the ground?
To determine the initial speed (also known as the initial velocity), we need to find the rate at which the armadillo is changing its vertical velocity. We can use the following equation:
v = u + at
where:
v = final velocity (which is 0, as the armadillo reaches its maximum height)
u = initial velocity (what we need to find)
a = acceleration (which is -9.8 m/s^2, taken as negative since the direction of motion is upwards)
t = time (0.190 s)
Since the armadillo starts from rest when it leaves the ground, the initial velocity is 0 m/s. Therefore, the answer is:
u = 0 m/s
(b) What is its speed at the height of 0.596 m?
At the maximum height, the speed of the armadillo will be zero as it momentarily stops before falling back down. So, the answer is:
v = 0 m/s
(c) How much higher does it go?
To determine the height reached by the armadillo, we can use the following equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s at the maximum height)
u = initial velocity (which is 0 m/s)
a = acceleration (which is -9.8 m/s^2, taken as negative since the direction of motion is upwards)
s = displacement (the height reached, which we need to find)
Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)
Substituting the values, we get:
s = (0^2 - 0^2) / (2 * -9.8)
s = 0 / (2 * -9.8)
s = 0
Therefore, the armadillo does not go any higher than 0.596 m.