Ch 03: 2D Motion (Projectile Motion)WorksheetSee all chapters
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Ch 01: Units & Vectors
Ch 02: 1D Motion (Kinematics)
Ch 03: 2D Motion (Projectile Motion)
Ch 04: Intro to Forces (Dynamics)
Ch 05: Friction, Inclines, Systems
Ch 06: Centripetal Forces & Gravitation
Ch 07: Work & Energy
Ch 08: Conservation of Energy
Ch 09: Momentum & Impulse
Ch 10: Rotational Kinematics
Ch 11: Rotational Inertia & Energy
Ch 12: Torque & Rotational Dynamics
Ch 13: Rotational Equilibrium
Ch 14: Angular Momentum
Ch 15: Periodic Motion (NEW)
Ch 15: Periodic Motion (Oscillations)
Ch 16: Waves & Sound
Ch 17: Fluid Mechanics
Ch 18: Heat and Temperature
Ch 19: Kinetic Theory of Ideal Gasses
Ch 20: The First Law of Thermodynamics
Ch 21: The Second Law of Thermodynamics
Ch 22: Electric Force & Field; Gauss' Law
Ch 23: Electric Potential
Ch 24: Capacitors & Dielectrics
Ch 25: Resistors & DC Circuits
Ch 26: Magnetic Fields and Forces
Ch 27: Sources of Magnetic Field
Ch 28: Induction and Inductance
Ch 29: Alternating Current
Ch 30: Electromagnetic Waves
Ch 31: Geometric Optics
Ch 32: Wave Optics
Ch 34: Special Relativity
Ch 35: Particle-Wave Duality
Ch 36: Atomic Structure
Ch 37: Nuclear Physics
Ch 38: Quantum Mechanics

Solution: In the figure, a stone is projected at a cliff of height  h with an initial speed of 42.0 m/s directed at angle θ0 = 60.0° above the horizontal. The stone strikes at C, 5.50 s after launching. Find (a

Problem

In the figure, a stone is projected at a cliff of height  h with an initial speed of 42.0 m/s directed at angle θ0 = 60.0° above the horizontal. The stone strikes at C, 5.50 s after launching. Find 

(a) the height h of the cliff, 

(b) the speed of the stone just before impact at C, and 

(c) the maximum height H reached above the ground.

Solution

In this problem, we’re asked to calculate the height of the cliff the stone lands on, the final speed of the stone, and the maximum height the stone reaches during its motion.

This is a projectile motion problem with a positive launch angle. For projectile motion problems, we follow the following simple steps:

  1. Draw diagram, axes.
  2. Develop equations to describe the various intervals.
  3. Solve each of the target variables. 

Step 1: Draw diagram, axes.

For projectiles with a positive launch angle, the motion has two parts: AB, from launch to peak, and BC, from peak to when it lands. We use AC for the full motion.

We'll consider the x and y speed components separately. Therefore, to resolve v0 to v0x and v0y, we apply trigonometry

vx=v0x=v0Cosθ        =(42.0m/s)Cos (60.0)

vx = 21.0 m/s

The horizontal speed (x-component of v0) is constant since there is no horizontal acceleration.

voy = v0Sinθ = (42.0 m/s) Sin(60.0°)

v0y = 36.4 m/s

The acceleration of the stone before reaching maximum height causes a decrease in vertical speed, vy while the acceleration after the maximum height increases the vertical speed, vy, downwards.

We will also use a standard coordinate system where upward motion is positive y and downward motion is negative y.  Also, the stone moves to the right in the positive x-direction.  

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