Projectile Activity 4 Solutions
1
| ) Use the complying with 2 kinematics formulas to reveal that the above projectile complies with an allegorical course. | |
| Δy = | |
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v i Δt + 1/2g Δt2 |
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Δx = vaveΔt |
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| v | |
| H = Δx/ Δt = vRcosθ | |
| Δt = Δx/(vRcosθ) | |
| Δy | |
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= viΔt + 1/2gΔt2 = vRsinθΔt-1/2gΔt 2 Replacing Δt=Δx/(v R cosθ)returns: Δy=v R sinθ –Δx/(v R cosθ)-1/2g – Δx 2/(v R | |
| 2 cos 2 θ)Δy= | |
| v R sinθ – Δx/(v R cosθ)-1/2g– Δx 2/(v R 2 cos 2 θ )Δy=tanθ – Δx | |
| – | |
| g – Δx 2/(2v R 2 cos 2 θ )Δy=a x-b x 2 where a=tanθas well as b=-g/(2vR2 cos 2 θ)are constants in the lack of air resistance.2)Aspace-flight projectile isreleased in an allegorical trajectory.An astronaut in this | |
| method pill | |
| really feels insubstantial.(a)Why? In the lack of air resistance, a projectile remains in cost-free autumn due to the fact that the only | |
| his upright velocity he | is not absolutely lightweight. If he was genuinely lightweight, after that as an outcome of his launch, he would certainly remain to relocate a straight line at consistent rate forever in the lack of air resistance or any kind of outside pressure. 4 Physlets Residence PowerPoint Projectile Motion Calculator Projectile Motion Solver
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