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| Projectile Motion |
- A body in projectile motion will move in two dimensions simultaneously. Bodies that are in projectile motion are resolved(divided) into two components. A vertical component(Vy) and an horizontal component(Vx). These components however are dealt with simultaneously.
Please refer to the diagram below showing projectile motion.
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| Projectile motion |
Formulas associated with projectile motion include:
- Vy = VsinA
- Vx = VcosA
- V = Sqrt [Vy^2 + Vx^2]
- TanA = Vy / Vx => A = Tan^-1 A (Vy / Vx)
Example calculation involving projectile motion:
- A body is projected with a velocity of 200m/s at and angle of 30 °C to the horizontal.
- Calculate the time taken to reach its maximum height.
- Calculate its velocity after 16 seconds.
Answers:
- Finding the time for maximum height.
V = 200m/s
Acceleration(a) = -10 m/s2
Vy = 0 m/s
Therefore Uy = VsinA [ Note: Uy represents initial velocity]
Uy = 200 x Sin 30
Uy = 100 m/s
Vy = Uy = at
Vy - Uy = at
a
0 - 100 = t
- 10
10 sec = t
Therefore Maximum Time (t) = 10 seconds
- Finding the velocity after 16 seconds.
Vx = V cos A
Vx = 200 x Cos 30
Vx = 173.2 m/s
Vy = Uy + at
Vy = 100 - 10 (16)
Vy = -60 m/s
Therefore V = √[Vy^2 + Vx^2]
= √[(-60)^2 + (173.2)^2]
= √33598
= 183.30 m/s