Question

A fox locates its prey, usually a mouse, under the snow by slight sounds the rodents make. The fox then leaps straight into the air and burrows its nose into the snow to catch its next meal. In your calculations ignore the effects of air resistance. 1) If a fox jumps to a height of 81.0 cm. Calculate the speed at which the fox leaves the snow. (Express your answer to three significant figures.)

Answer 1

The speed at which the fox leaves the snow is approximately 4.52 m/s.

Step-by-step explanation:

To calculate the speed at which the fox leaves the snow, we can use the principle of conservation of energy. Since the fox jumps straight up, its initial vertical velocity is zero. The final velocity can be calculated using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. In this case, the displacement is the height jumped, which is 81.0 cm or 0.81 m. The acceleration is equal to g, the acceleration due to gravity (approximately 9.8 m/s^2). Solving for v, we find that the speed at which the fox leaves the snow is approximately 4.52 m/s.

Answer 2

The fox leaves the snow at 3.99 m/s

Step-by-step explanation:

Hi there!

The equation of height and velocity of the fox are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the fox at a time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s considering the upward direction as positive)

v = velocity of the fox at a time t.

We know that at the maximum height of the fox, its velocity is zero, so using the equation of velocity we can obtain an expression of v0 in function of t:

v = v0 + g · t

At the maximum height, v = 0

0 = v0 + g · t

Solving for v0:

-g · t = v0

We know the maximum height of the fox, 0.810 m. So, using the equation of height and replacing v0 by (-g · t), we can obtain the time at which the fox is at the maximum height and then calculate the initial velocity:

h = h0 + v0 · t + 1/2 · g · t²

When t is the time at which the fox is at the maximum height, h = 0.810 m and v0 = (-g · t). Let´s consider the ground as the origin of the frame of reference so that h0 = 0.

0.810 m = (-g · t) · t + 1/2 · g · t²

0.810 m = -g · t² + 1/2 · g · t²

0.810 m = - 1/2 · g · t²

t² = -2 · 0.810 m / -9.81 m/s²

t = 0.406 s

And the initial velocity will be:

v0 = -g · t

v0 = -(-9.81 m/s²) · 0.406 s

v0 = 3.99 m/s

The fox leaves the snow at 3.99 m/s