Question

problem 8: a mountain climber must jump over a crevasse. she leaps horizontally at a speed $v_0 = 5 m/s$. what must the height difference between the two sides be if the final angle of motion upon landing by the climber is $55^{circ}$ below the horizontal.

Explanation:

Step1: Analyze horizontal - vertical velocities

The initial horizontal velocity $v_{0x}=v_0 = 5\ m/s$ (constant in projectile motion as there is no horizontal acceleration). Let the final vertical velocity be $v_y$. We know that $\tan\theta=\frac{v_y}{v_x}$, where $\theta = 55^{\circ}$ and $v_x = v_{0x}=5\ m/s$.
$\tan(55^{\circ})=\frac{v_y}{5}$

Step2: Solve for final vertical velocity

Since $\tan(55^{\circ})\approx1.4281$, then $v_y = 5\times\tan(55^{\circ})\approx5\times1.4281 = 7.1405\ m/s$.

Step3: Use kinematic equation for vertical motion

The kinematic equation $v_y^{2}=v_{0y}^{2}+2gh$ is used. The initial vertical velocity $v_{0y} = 0\ m/s$ (leaps horizontally). So $v_y^{2}=2gh$. We want to find the height difference $h$.
$h=\frac{v_y^{2}}{2g}$

Step4: Calculate the height difference

Taking $g = 9.8\ m/s^{2}$, we substitute $v_y\approx7.1405\ m/s$ into the formula:
$h=\frac{(7.1405)^{2}}{2\times9.8}=\frac{50.987}{19.6}\approx2.6\ m$

Answer:

$2.6\ m$