How to Calculate Projectile Motion for a Thrown Stone

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SUMMARY

The discussion focuses on calculating the projectile motion of a stone thrown horizontally from a height of 38 meters with an initial speed of 7.6 m/s. The time to impact can be determined using the equation for vertical motion, yielding approximately 2.8 seconds. The final velocity at impact combines both horizontal and vertical components, with the vertical velocity calculated using the equation v^2 = u^2 + 2as. The angle of impact below the horizontal is derived from the velocity components, forming a right triangle.

PREREQUISITES
  • Understanding of basic physics concepts, specifically projectile motion
  • Familiarity with kinematic equations, including v^2 = u^2 + 2as
  • Knowledge of gravitational acceleration, specifically g = -9.8 m/s²
  • Ability to calculate angles using trigonometric functions
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to apply kinematic equations in different scenarios
  • Explore the effects of air resistance on projectile motion
  • Practice solving problems involving two-dimensional motion
USEFUL FOR

This discussion is beneficial for physics students, educators teaching projectile motion, and anyone interested in understanding the principles of motion in a gravitational field.

anna_chem
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Homework Statement


a student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 7.6 m/s. The cliff is 38 m above a flat, horizontal beach. Find how long it takes to hit the beach, the velocity at impact, and at what angle below the horizontal the stone is when it lands.


Homework Equations


y=Vo*sin(initial angle)*t - .5gt^2
Vox= 7.6 m/s y-yo=38 and g=-9.8m/s^2

The Attempt at a Solution


I'm having a hard time getting started, since I only know Vx, the height, and g. I believe the initial angle would be 0 degrees since the student is throwing the stone horizontally and not vertically.
 
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Well, since the problem states that he is throwing the stone horizontally with no reference to vertical motion, the time it takes to hit the ground will be the same as a rock that has been dropped from that height. I myself have only started taking Physics I now, so I am unsure about the other parts of the problem.
 
To work out at what time it hits the ground you can just use v^2 = u^2 + 2as, to get it's vertical velocity when it hits the ground, and then use v = u +at to get the time. This is because it's vertical velocity is unaffected by it's horizontal one.

Likewise it's horizontal velocity remains unchanged throughtout the flight as there is no air resistance. The velocity when it hit's the ground is found by combining the vertical and horizontal velocities it has when it hits the ground. The velocity components form a triangle which you can use to get the angle.

I would try drawing it out, will help you.
 
Last edited:

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