A daredevil drove his motorcycle up an incline at 40 degree angle with the horizontal and drove off the end of the incline at 22 m/s, 12 m above the ground. How fast was him (magnitude of total velocity) when he landed?
Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
![**Physics Problem: Projectile Motion**
A daredevil drove his motorcycle up an incline at a 40-degree angle with the horizontal and drove off the end of the incline at 22 m/s, 12 meters above the ground. How fast was he (magnitude of total velocity) when he landed?
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**Solution Explanation:**
To solve this problem, we need to apply the principles of projectile motion. Here's a step-by-step guide on how to calculate the magnitude of the velocity when the daredevil lands.
1. **Decompose the Initial Velocity:**
- **Horizontal Component (Vx):**
\[
Vx = V \cos(\theta) = 22 \, m/s \cdot \cos(40^{\circ})
\]
\[
Vx \approx 16.85 \, m/s
\]
- **Vertical Component (Vy):**
\[
Vy = V \sin(\theta) = 22 \, m/s \cdot \sin(40^{\circ})
\]
\[
Vy \approx 14.14 \, m/s
\]
2. **Calculate the Time of Flight (t):**
- Using the equation for vertical displacement under gravity:
\[
y = Vy \cdot t + \frac{1}{2}gt^2
\]
Here, \( y = -12 \, m \) (since the daredevil lands 12 meters below the takeoff point), and \( g = 9.8 \, m/s^2 \). Rearrange and solve the quadratic equation for \( t \):
\[
-12 = 14.14 \, t - 4.9 \, t^2
\]
Solving this quadratic equation gives \( t \approx 3.67 \, s \).
3. **Calculate the Final Vertical Velocity (Vy_final) just before landing:**
- Using the equation of motion:
\[
Vy_{final} = Vy + gt
\]
\[
Vy_{final} = 14.14 - 9.8 \times 3.67
\]
\[
Vy_{final} \approx -21.76 \, m/s
\]
4. **Calculate the Magnitude of the Total Velocity:**
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd2932a05-99f3-41c1-989e-7dbca32c3377%2F94b120e9-2114-4cb7-b688-8ced12e303ee%2F3wd2nhb_processed.png&w=3840&q=75)

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