Physics Problem: Projectile Motion

Physics Problem: Projectile Motion

#Physics #Kinematics #ProjectileMotion #ResnickHallidayWalker #TajSirPhysics #ProblemSolving

Problem Statement

Course: Physics I
Instructor: Taj Sir
Source: Resnick Halliday Walker, Chapter 4

A projectile is launched from ground level with an initial speed of 50 m/s at an angle of 37° above the horizontal. It strikes a target above the ground 3.0 seconds later.

(a) What are the x and y coordinates of the target?
(b) What is the speed of the projectile just before it strikes the target?

Projectile Motion Diagram

37°
Target: (120m, 45.9m)
Launch Point (0,0)
Parabolic Trajectory
Target Position

Visual representation showing the parabolic trajectory of the projectile

Solution

Step 1: Break Initial Velocity into Components

The initial velocity can be resolved into horizontal and vertical components:

\[ v_{0x} = v_0 \cos \theta \] \[ v_{0y} = v_0 \sin \theta \]

Where:
- \( v_0 = 50 \, \text{m/s} \)
- \( \theta = 37^\circ \)

Using trigonometric values:
- \( \cos 37^\circ \approx 0.8 \)
- \( \sin 37^\circ \approx 0.6 \)

\[ v_{0x} = 50 \times 0.8 = 40 \, \text{m/s} \] \[ v_{0y} = 50 \times 0.6 = 30 \, \text{m/s} \]

Step 2: Calculate Horizontal Position (x-coordinate)

In projectile motion, horizontal velocity remains constant (neglecting air resistance):

\[ x = v_{0x} \cdot t \]

Where:
- \( v_{0x} = 40 \, \text{m/s} \)
- \( t = 3.0 \, \text{s} \)

\[ x = 40 \times 3.0 = 120 \, \text{m} \]

Step 3: Calculate Vertical Position (y-coordinate)

The vertical position is given by:

\[ y = v_{0y} \cdot t - \frac{1}{2} g t^2 \]

Where:
- \( v_{0y} = 30 \, \text{m/s} \)
- \( t = 3.0 \, \text{s} \)
- \( g = 9.8 \, \text{m/s}^2 \)

\[ y = (30 \times 3.0) - \frac{1}{2} (9.8) (3.0)^2 \] \[ y = 90 - (4.9 \times 9) \] \[ y = 90 - 44.1 = 45.9 \, \text{m} \]
Answer to part (a): \( \boxed{(120\ \text{m},\ 45.9\ \text{m})} \)

Step 4: Calculate Final Speed

The horizontal velocity remains constant:

\[ v_x = v_{0x} = 40 \, \text{m/s} \]

The vertical velocity at time t is given by:

\[ v_y = v_{0y} - g t \]
\[ v_y = 30 - (9.8 \times 3.0) = 30 - 29.4 = 0.6 \, \text{m/s} \]

The final speed is the magnitude of the velocity vector:

\[ v = \sqrt{v_x^2 + v_y^2} \]
\[ v = \sqrt{(40)^2 + (0.6)^2} = \sqrt{1600 + 0.36} = \sqrt{1600.36} \approx 40.0045 \, \text{m/s} \]
Answer to part (b): \( \boxed{40.0\ \text{m/s}} \)

Final Answers

(a) Coordinates: (120 m, 45.9 m)
(b) Final speed: 40.0 m/s

Key Concepts

  • Projectile Motion: Motion under constant acceleration (gravity)
  • Independence of Motion: Horizontal and vertical motions are independent
  • Horizontal Velocity: Remains constant in projectile motion (no horizontal acceleration)
  • Vertical Velocity: Changes linearly with time due to constant acceleration g
  • Trajectory: Parabolic path when air resistance is negligible

This problem demonstrates the application of kinematic equations to two-dimensional motion, a fundamental concept in mechanics with applications in sports, engineering, and ballistics.

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