Question

A track star in the long jump goes into the jump at 12 m/s and launches herself at 20.0 degrees above the horizontal. What is the magnitude of her horizontal displacement?

a. 4.6 m
b. 9.2 m
c. 13 m
d. 15 m

Answer 1

b. 9.2m

Step-by-step explanation:

Vertical component of velocity:

12sin20

a = -9.8

t for max height

0 = 12sin20 - 9.8t

t = 0.4188001755

Time of the journey:

2t = 0.837600351

Horizontal component of velocity:

12cos20

Horizontal displacement:

(12cos20)(0.837600351)

9.445042428 m

Answer 2

`\large \boxed{\text{b. 9.4 m}}`

Step-by-step explanation:

1. Break down the velocity into its horizontal and vertical components

`\cos 20^(\circ) = \frac{v_{\text{h}}}{12} = 0.9397\\\\v_{\text{h}} = \text{11.3 m/s}`

`\sin 20 = \frac{v_{\text{v}}}{12} = 0.3420\\\\v_{\text{v}} = \text{4.10 m/s}`

2. Calculate the time of flight

Use the vertical component of velocity to calculate the time to the height of the jump.

`v = gt\\t = (v)/(g) = \frac{\text{4.10 m$\cdot$ s}^(-1)}{\text{9.807 m$\cdot$ s}^(-2)}= \text{0.419 s}`

It will take the same time to reach the ground.

Thus,

Time of flight = 2t = 2 × 0.419 s = 0.837 s

3. Calculate the horizontal distance

s = vt = 11.3 m·s⁻¹ × 0.837 s = 9.4 m

`\text{Her horizontal displacement is $\large \boxed{\textbf{9.4 m}}$}`