To solve this problem, we can break the squirrel’s motion into horizontal and vertical components, treating it as a projectile motion problem.

Given dataTo solve this problem, we can break the squirrel’s motion into horizontal and vertical components, treating it as a projectile motion problem.

Given

• Initial speed, $v_0 = 99.0 \, \text{m/s}$
• Angle of projection, $\theta = 45^\circ$
• Height of the building, $h = 25.0 \, \text{m}$
• Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

A) Time in the Air

We can use the kinematic equation for vertical motion to calculate the time the squirrel spends in the air. The vertical displacement $\Delta y$ between the initial position (ground) and the final position (roof of the building) is -25.0 m (since it is lower than its maximum height).

The vertical motion is described by: $\Delta y = v_{0y} t - \frac{1}{2} g t^2$ where: $v_{0y} = v_0 \sin \theta$ Substituting known values: $\Delta y = -25.0 \, \text{m}, \, v_0 = 99.0 \, \text{m/s}, \, \theta = 45^\circ, \, g = 9.81 \, \text{m/s}^2$

This gives us a quadratic equation to solve for time $t$.

B) Maximum Height Above the Ground

The maximum height occurs when the vertical velocity becomes zero. The equation for maximum height is: $h_{\text{max}} = \frac{(v_{0y})^2}{2g}$

We will first calculate $v_{0y}$ and then use it to find $h_{\text{max}}$.

C) Horizontal Distance

The horizontal distance traveled is given by: $x = v_{0x} \cdot t$ where: $v_{0x} = v_0 \cos \theta$ We will use the time of flight (calculated in part A) to determine the horizontal distance.

Let me calculate each part step by step.