Answer to Question #257193 in Physics for temmy

Question

 A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

Answer 1

(a) Let's first find the time that the long jumper takes to reach the peak height:

$$v_y = v_{0y}-gt,$$ $$0=v_0sin\theta-gt,$$ $$t=\dfrac{v_0sin\theta}{g}.$$

Then, we can find the time of flight as follows:

$$t_{flight}=2t=\dfrac{2v_0sin\theta}{g},$$ $$t_{flight}=\dfrac{2\times12\ \dfrac{m}{s}\times sin28^{\circ}}{9.8\ \dfrac{m}{s^2}}=1.15\ s.$$

(b) We can find the horizontal distance from the kinematic equation:

$$x=v_0tcos\theta=12\ \dfrac{m}{s}\times1.15\ s\times cos28^{\circ}=12.18\ m.$$

(c) We can find the peak height of the long-jumper from the kinematic equation:

$$y_{max}=v_0tsin\theta-\dfrac{1}{2}gt^2.$$

Substituting $t$ into the second equation, we get:

$$y_{max}=\dfrac{v_0^2sin^2\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2=\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=\dfrac{(12\ \dfrac{m}{s})^2\times sin^228^{\circ}}{2\times9.8\ \dfrac{m}{s^2}}=1.62\ m.$$