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)*

Question

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)