Question

A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0° above the horizontal. How long is she in the air before returning to Earth? (g = 10 m/s2) t = (2*Vy)/g *(Remember find components first using sin and cos)*
a
0.82 s
b
0.42 s
c
1.2 s
d
1.5 s
Question 18 (1 point)
A model rocket flies horizontally off the edge of the cliff at a velocity of 50.0 m/s. If the canyon below is 100.0 m deep, how far from the edge of the cliff does the model rocket land?
t = mc012-1.jpg x = t * v
a
448 m
b
224 m
c
337 m
d
112 m

Answer 1

Let's solve each problem one by one.

### Problem 1: Long Jump

1. **Find the vertical component of velocity (Vy)**:

\[
Vy = V \cdot \sin(\theta)
\]
Where:
- \( V = 12 \, \text{m/s} \)
- \( \theta = 20^\circ \)

\( Vy = 12 \cdot \sin(20^\circ) \)

To calculate \( \sin(20^\circ) \):
- \( \sin(20^\circ) \approx 0.342 \)

Thus:
\[
Vy \approx 12 \cdot 0.342 \approx 4.104 \, \text{m/s}
\]

2. **Time in the air (t)**:

The formula for the total time of flight when it returns to the same vertical level is:
\[
t = \frac{2 \cdot Vy}{g}
\]
Where \( g = 10 \, \text{m/s}^2 \):

\[
t = \frac{2 \cdot 4.104}{10} \approx \frac{8.208}{10} \approx 0.8208 \, \text{s}
\]

So the answer is approximately **0.82 s** (Option a).

### Problem 2: Model Rocket

1. **Time to fall to the ground (t)**:

Using the formula for free fall:
\[
h = \frac{1}{2}gt^2
\]
Where \( h = 100.0 \, m \) and \( g = 10 \, m/s^2 \):

Rearranging for \( t \):
\[
100 = \frac{1}{2} \cdot 10 \cdot t^2
\]
\[
100 = 5t^2
\]
\[
t^2 = 20
\]
\[
t = \sqrt{20} \approx 4.47 \, \text{s}
\]

2. **Horizontal distance (x)**:

The horizontal distance traveled can be calculated using:
\[
x = t \cdot v
\]
Where \( v = 50.0 \, \text{m/s} \):

\[
x = 4.47 \cdot 50.0 \approx 223.5 \, \text{m}
\]

So the closest answer is approximately **224 m** (Option b).

### Summary of Answers:
- Long jump: **0.82 s** (Option a)
- Model rocket: **224 m** (Option b)