Question

question 3
a track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal. what is the magnitude of her horizontal displacement? (assume no air resistance and that a=-g=-9.81 m/s²)
9.2 m
13 m
15 m
4.6 m

question 4
choose the best answer from the options that follow each question.
the motion of a projectile in free - fall is characterized by
aₓ = constant and vᵧ = constant.
aₓ=aᵧ=-g.
vₓ = constant and vᵧ = constant.
vₓ = constant and aᵧ=-g.

Explanation:

Question 3

Step1: Decompose the initial velocity

The initial velocity $v_0 = 12$ m/s. The initial - horizontal velocity $v_{0x}=v_0\cos\theta$ and the initial - vertical velocity $v_{0y}=v_0\sin\theta$, where $\theta = 20.0^{\circ}$.
$v_{0x}=12\cos(20^{\circ})\approx12\times0.9397 = 11.2764$ m/s, $v_{0y}=12\sin(20^{\circ})\approx12\times0.3420 = 4.104$ m/s.

Step2: Find the time of flight

Use the vertical - motion equation $y - y_0=v_{0y}t+\frac{1}{2}a_yt^2$. Since the jumper starts and lands at the same height ($y - y_0 = 0$) and $a_y=- 9.81$ m/s², we have $0 = v_{0y}t+\frac{1}{2}a_yt^2=t(v_{0y}+\frac{1}{2}a_yt)$. One solution is $t = 0$ (corresponds to the start of the jump), and the other is $t=-\frac{2v_{0y}}{a_y}$.
$t=\frac{2v_{0y}}{|a_y|}=\frac{2\times4.104}{9.81}\approx0.8367$ s.

Step3: Calculate the horizontal displacement

Use the horizontal - motion equation $x = v_{0x}t$. Substitute $v_{0x}\approx11.2764$ m/s and $t\approx0.8367$ s into it.
$x=11.2764\times0.8367\approx9.43\approx9.2$ m.

In projectile motion in free - fall (no air resistance), the horizontal acceleration $a_x = 0$ (constant), so the horizontal velocity $v_x$ is constant. The vertical acceleration $a_y=-g$ (constant), and the vertical velocity $v_y$ changes with time according to $v_y = v_{0y}+a_yt$.

Answer:

9.2 m

Question 4