Find the initial velocity of a projectile

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SUMMARY

The discussion focuses on calculating the initial velocity (v0) of a long jumper using projectile motion principles. The range equation, x = v^2/(g sin(2θ)), was applied, where d = 7.87 m, g = 9.8 m/s², and θ = 21.4°. The user attempted to solve for velocity but failed to account for the difference in initial and final heights, leading to an incorrect application of the range equation. The correct approach requires adjustments for varying heights in projectile motion calculations.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with the range equation for projectile motion
  • Basic knowledge of trigonometric functions, specifically sine
  • Ability to manipulate algebraic equations for solving physics problems
NEXT STEPS
  • Study the derivation and application of the range equation for projectile motion with varying heights
  • Learn about the kinematic equations for projectile motion
  • Explore the impact of launch angle on projectile trajectories
  • Investigate the effects of initial and final heights on projectile motion calculations
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This discussion is beneficial for physics students, educators, and anyone involved in sports science or engineering, particularly those focusing on projectile motion and its applications in athletics.

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Homework Statement


So it's based off of a track&field long jumper. Calculate the initial speed, v0, of a long-jumper where d=7.87 m. Assume a take-off angle of 21.4°, an initial height of the centre-of-mass of 0.920 m, and a final height of 0.460 m.

So I use the range equation for projectile motion and rearrange it for velocity but it doesn't seem to work...


can someone correct me?


Homework Equations


x = v^2/gsin2(theta) where x is the range. i rearranged for velocity.


The Attempt at a Solution


x = v^2/gsin2(theta)

v = sqrt(xgsin2(theta)
= sqrt(7.87m*9.8*sin(2*21.4)
= 7.24 m/s
 
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Your equation works when initial and final height is the same. Here it isn't
 

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