Kinematics: Solving 2 Projectile Flight Time Diff.

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SUMMARY

The discussion focuses on calculating the difference in flight times between two projectiles launched at identical speeds of 30 m/s at angles of 40 degrees and 50 degrees. The flight times are determined using the vertical components of the initial velocities, calculated as v_{y1}=30sin(40)=19.283 m/s and v_{y2}=30sin(50)=22.981 m/s. The resulting flight times are t_{1}=3.931 seconds and t_{2}=4.685 seconds, leading to a time difference of 0.754 seconds, rounded to three significant figures.

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  • Basic knowledge of gravitational acceleration (g = 9.81 m/s²)
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Kyoma
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I have trouble solving this question:

2 projectiles are launched with identical speeds of 30m/s and at angles 40 degrees and 50 degrees with the horizontal, respectively. The difference between the flight times of the 2 projectiles is?

The answer given to me is 0.754s, round up to 3 significent figures.
 
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Just treat it as two separate projectile problems.

Velocity is 30m/s so for the first, resolve the velocity horizontally and vertically at an angle of 40 degrees.
You can then calculate the flight time using just the vertical velocity (v = u + at).

Do the same for the second and subtract the results.
 
[tex]v_{y1}=30sin(40)=19.283, v_{y2}=30sin(50)=22.981[/tex]

The flying times given that [tex]g=9.81m/s^2[/tex] are [tex]t_{1}=\frac{2v_{y1}}{g}=3.931,t_{2}=\frac{2v_{y2}}{g}=4.685[/tex]
 

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