Projectile motion rocket problem

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SUMMARY

The projectile motion problem involves a rocket launched at 75.0 m/s at a 60.0-degree angle, aimed at an 11.0 m high wall located 27.0 m away. The initial vertical velocity (vy) is calculated as 65.0 m/s, and the horizontal velocity (vx) is 37.5 m/s. The time to reach the wall is determined to be 0.720 seconds, resulting in a vertical position of 49.3 m at that time. The discrepancy in the final answer, which should be 33.2 m above the wall, arises from the incorrect application of vertical acceleration; it should be -9.8 m/s² instead of +9.8 m/s².

PREREQUISITES
  • Understanding of projectile motion principles
  • Basic trigonometry for resolving velocity components
  • Knowledge of kinematic equations for motion analysis
  • Familiarity with gravitational acceleration (9.8 m/s²)
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  • Review kinematic equations for projectile motion
  • Practice problems involving angle of launch and height clearance
  • Explore the effects of air resistance on projectile motion
  • Learn about the trajectory analysis of projectiles in physics
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Homework Statement


A rocket is fired at at speed fo 75.0 m/s from ground level, at an angle of 60.0 deg. above the horizontal. The rocket is fired toward an 11.0 m high wall, shich is located 27.0 m away. The rocket attains its launch speed in a negligibly short period of time, after which its engines shut down and the rocket coasts. By how much does the rocket clear the top of the wall?


Homework Equations





The Attempt at a Solution


I got the initial vy by 75.0(sin(60))=65.0 m/s, and vx 75.0(cos(60))=37.5 m/s
27.0m=(37.5m/s)t + (1/2*0*t)
t=0.720s
y=65.0 m/s(0.720s) + 1/2(9.80 m/s^2)(0.720^2)
y=49.3m
49.3-11.0=38.3m above the wall
The answer is supposed to be 33.2m, I'm not sure where I went wrong.
 
Physics news on Phys.org
vertical acceleration = -9.8m/s^2 not 9.8m/s^2
 
thanks
 

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