Uncovering Projectile Motion Properties

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a projectile launched with an unknown initial velocity that lands 19.8 seconds later, 2710 meters horizontally and 463 meters vertically above its starting point. The vertical component of the initial velocity can be calculated using the equation y = y_0 + v_y t + 0.5 a t^2, leading to a vertical initial velocity of approximately 19.6 m/s. The horizontal component can be determined using the horizontal distance and time, resulting in a horizontal initial velocity of about 137.4 m/s. The maximum height above the launch point is found by determining the time when the vertical velocity is zero, and the final speed and angle upon impact can be calculated from the components of velocity at t = 19.8 seconds.

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  • Understanding of 2D kinematics equations
  • Knowledge of projectile motion concepts
  • Familiarity with basic physics principles, including acceleration due to gravity
  • Ability to solve quadratic equations
NEXT STEPS
  • Learn to apply 2D kinematics equations in different projectile motion scenarios
  • Explore the effects of air resistance on projectile motion
  • Study the derivation and application of the range formula for projectiles
  • Investigate the relationship between launch angle and maximum height in projectile motion
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Physics students, educators, and anyone interested in mastering projectile motion calculations and applications in real-world scenarios.

creativeone
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A projectile, fired with unknown initial velocity, lands 19.8 s later on the side of a hill, 2710 m away horizontally and 463 m vertically above its starting point. (Ignore any effects due to air resistance.)

(a) What is the vertical component of its initial velocity?
(b) What is the horizontal component of its initial velocity?
(c) What was its maximum height above its launch point?
(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?




Known 2D kinematics equations



3. How would you do this, I have no idea.
 
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It's just like the baseball problem you did.
Start with the y component.
y = y_0 + v_y t + \frac{1}{2} a t^2

Assuming y_0 = 0
y(t=19.8 s) = 463 m = v_y (19.8s) + \frac{1}{2} (-9.8 m/s) (19.8s)^2

Part b is similar.

Part c. Find the time when v_y = 0.

Part d. From the components of velocity at t = 19.8s, what is the speed? What is the direction?
 

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