Questions: A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its nose into the snow to catch its meal. If a fox jumps up to a height of 93 cm, calculate the speed v at which the fox leaves the snow and the amount of time t the fox is in the air. Ignore air resistance. v= m / s t= s

Solution Steps

To find the speed \( v \) at which the fox leaves the snow, we use the principle of conservation of energy. The potential energy at the maximum height is equal to the kinetic energy at the point of takeoff.

The potential energy at the maximum height \( h \) is given by: \[ PE = mgh \] where \( m \) is the mass of the fox, \( g \) is the acceleration due to gravity (9.81 m/s\(^2\)), and \( h \) is the height (0.93 m).

The kinetic energy at takeoff is given by: \[ KE = \frac{1}{2}mv^2 \]

Setting the potential energy equal to the kinetic energy: \[ mgh = \frac{1}{2}mv^2 \]

Solving for \( v \): \[ gh = \frac{1}{2}v^2 \] \[ v^2 = 2gh \] \[ v = \sqrt{2gh} \]

Substituting \( g = 9.81 \, \text{m/s}^2 \) and \( h = 0.93 \, \text{m} \): \[ v = \sqrt{2 \times 9.81 \times 0.93} \] \[ v = \sqrt{18.2346} \] \[ v \approx 4.270 \, \text{m/s} \]

The total time \( t \) the fox is in the air can be found by considering the time to reach the maximum height and the time to fall back down. Since the motion is symmetrical, the time to reach the maximum height is equal to the time to fall back down.

The time to reach the maximum height \( t_{\text{up}} \) can be found using the kinematic equation: \[ v = gt_{\text{up}} \] Solving for \( t_{\text{up}} \): \[ t_{\text{up}} = \frac{v}{g} \]

Substituting \( v = 4.270 \, \text{m/s} \) and \( g = 9.81 \, \text{m/s}^2 \): \[ t_{\text{up}} = \frac{4.270}{9.81} \] \[ t_{\text{up}} \approx 0.4351 \, \text{s} \]

The total time in the air \( t \) is twice the time to reach the maximum height: \[ t = 2 \times t_{\text{up}} \] \[ t = 2 \times 0.4351 \] \[ t \approx 0.8702 \, \text{s} \]

Final Answer