How high will the projectile be at 5 km downrange?

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SUMMARY

The projectile is launched with an initial speed of 500 m/s at a 45° angle. The flight time is calculated to be 72.2 seconds, resulting in a horizontal distance of 25,526.6 meters. To determine the height of the projectile at 5 km downrange, the formula r(t) = v₀cosθ(t) + (v₀sinθ(t) - 0.5gt²) is applicable. The time calculated earlier (72.2 seconds) is not directly usable for the 5 km distance; a separate calculation for the time to reach 5 km is necessary.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions, specifically sine and cosine
  • Knowledge of kinematic equations
  • Basic grasp of gravitational acceleration (9.8 m/s²)
NEXT STEPS
  • Calculate the time it takes for the projectile to reach 5 km downrange using the horizontal motion equation.
  • Apply the vertical motion equation to find the height at the calculated time.
  • Review the derivation of projectile motion equations for better understanding.
  • Explore the effects of varying launch angles on projectile height and distance.
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to enhance their teaching methods in these topics.

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Homework Statement



A projectile is fired with an initial speed of 500 m/s at an angle of elevation of 45°. When and how far away will the projectile strike? How high overhead will the projectile be when it is 5 km downrange?

Homework Equations





The Attempt at a Solution


Flight time: t=\frac{2(500)sin45°}{9.8}
t=72.2 sec
How far: x=500cos45°(72.2)
x=25526.6 m

I'm not entirely sure how to figure out the second part though. I was thinking about using this formula: r(t)=vocosθ(t)+(vosinθ(t)-\frac{1}{2}gt2)
r(t)=x+(y-\frac{1}{2}gt2)
Am I on the right track with this? Do I use the t I calculated earlier for this equation?
 
Last edited:
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How long will it take to be that far down range?
 

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