Height of bullet at top of trajectory

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SUMMARY

The discussion focuses on calculating the height a bullet reaches when fired straight up at 1000 mph and determining the angle required for the bullet to have a speed of 100 mph at the peak of its trajectory. For the first part, the kinematic equation \( V^2 = Vo^2 + 2ax \) is applicable, where \( V \) is the final velocity (0 at the peak), \( Vo \) is the initial velocity (1000 mph), \( a \) is the acceleration due to gravity, and \( x \) is the height. The second part involves analyzing projectile motion to find the necessary launch angle and the corresponding height at which the bullet's speed reduces to 100 mph.

PREREQUISITES
  • Understanding of kinematic equations, specifically \( V^2 = Vo^2 + 2ax \)
  • Basic principles of projectile motion
  • Knowledge of gravitational acceleration (approximately 32.2 ft/s² or 9.81 m/s²)
  • Familiarity with converting units between mph and other speed measures
NEXT STEPS
  • Calculate the maximum height using the kinematic equation for the first part of the problem
  • Explore projectile motion concepts to determine the launch angle for the second part
  • Learn about energy conservation principles in projectile motion
  • Investigate the effects of air resistance on projectile trajectories
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion and its applications in real-world scenarios.

calnpals
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Ok guys, I'm stuck on this question I was wondering if you could help me out.

Its actually in 2 parts (2nd part harder than 1st part).

1st part: At what height will a bullet reach if fired at 1000mph straight up (excluding all other forces except gravity)

2nd part: At what angle would a bullet that is shot at 1000mph have to be shot at to have a speed of 100mph at the height of its trajectory, and what height would it reach?

Thanks
 
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It would be better if you showed some of your work.
Anyway, for the first part you just need a basic kinimatics equation
( V^2 = Vo^2 + 2ax) or you could use energy.
And for the second part, at the height of the trajectory the only speed is along the x-axis and the speed along that axis is the same throughout the whole trip. Does that make it any easier?
 
calnpals said:
Ok guys, I'm stuck on this question I was wondering if you could help me out.

Its actually in 2 parts (2nd part harder than 1st part).

1st part: At what height will a bullet reach if fired at 1000mph straight up (excluding all other forces except gravity)

2nd part: At what angle would a bullet that is shot at 1000mph have to be shot at to have a speed of 100mph at the height of its trajectory, and what height would it reach?

Thanks
How do you start ? What formula's will you use and more importantly why ?

marlon
 

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