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Projectile Motion: Questions to Ponder

Projectile Motion: Questions to Ponder. • What are the three factors that will determine how far an object will travel when thrown? What are the two things a vector must show? • Why does a baseball pitcher throw from a mound?. Analyzing Projectile Motion.

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Projectile Motion: Questions to Ponder

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  1. Projectile Motion: Questions to Ponder • What are the three factors that will determine how far an object will travel when thrown? • What are the two things a vector must show? • Why does a baseball pitcher throw from a mound?

  2. Analyzing Projectile Motion Projectile motion is a motion resulting from an initial velocity and subject only to the acceleration of gravity.

  3. Analyzing Projectile Motion 3 seconds down 3 seconds up The time it takes for an object to reach its apex is the same as the time it takes to go from the apex to the ground.

  4. Analyzing Projectile Motion 3 seconds down 3 seconds up The initial velocity is the same as the final velocity. Initial velocity is a POSITIVE number; final velocity is NEGATIVE.

  5. Projectile MotionLaws of Constant Acceleration Both dimensions follow the motion equations of previous studies Vf=Vi + at d=1/2(Vf + Vi)t d=Vit + 1/2at2 Vf2 = Vi2 + 2ad Vertical Component Horizontal Component

  6. Suggestions for Solving PM Problems 1. Visualize the problem – make a drawing 2. Decide what your coordinate system is: which directions are +,which are – 3. Remember the time variable, t, is what binds the x motion to the y motion (same value for the part of the motion). 4. Write down all your known and implied values in an organized way 5. Verify that the given information contains values for at least 3 of the kinematic variables. Do this for both x and y motions. Then select the appropriate equation(s). 6. You may want to divide the motion into segments. Remember the final velocity for one segment is the initial velocity for the next segment. 7. Remember that a problem may have two possible answers. Try to visualize the different physical situations to which the answers correspond.

  7. Example – Launched at Zero Angle A plane flying at an altitude of 1050 m and a speed of 115 m/s plans to drop a bunker-buster bomb on OBL’s hideout. How long will it take the bomb to reach its target? How fast will the bomb be traveling upon impact (net)? How far (surface distance) from the target should the bomb be released? 115 m/s 1050 m

  8. Example – Launched at Zero Angle A plane flying at an altitude of 1050 m and a speed of 115 m/s plans to drop a bunker-buster bomb on OBL’s hideout. How long will it take the bomb to reach its target? Fall time is independent of the horizontal speed Y-dimension values: Vi = 0 a= 9.80 m/s2 d = 1050 m t = ? apply: d=Vit + 1/2at2 , 2d/a = t2 , √2d/a = t t = √2(1050m)9.80m/s2 t = 14.6s 115 m/s 1050 m

  9. Example – Launched at Zero Angle A plane flying at an altitude of 1050 m and a speed of 115 m/s plans to drop a bunker-buster bomb on OBL’s hideout. How fast will the bomb be traveling upon impact (net)? 115 m/s ? m/s net The vertical velocity upon mpact is independent of the horizontal speed Vi = 0 a= 9.80 m/s2 d = 1050 m Vf = ? Apply: Vf2 = Vi2 + 2ad , Vf = √Vi2 + 2ad Vf = √2(9.80m/s2)1050m Vf = 144 m/s 115 m/s 1050 m

  10. Example – Launched at Zero Angle A plane flying at an altitude of 1050 m and a speed of 115 m/s plans to drop a bunker-buster bomb on OBL’s hideout. How fast will the bomb be traveling upon impact (net)? 115 m/s 144 m/s net Use the Pythagorean theorem to resolve two vectors occurring at right angles. C2 = A2+ B2 C = √A2+ B2 C = √(144 m/s)2 + (115 m/s)2 C = 184 115 m/s 1050 m

  11. Example – Launched at Zero Angle A plane flying at an altitude of 1050 m and a speed of 115 m/s plans to drop a bunker-buster bomb on OBL’s hideout. How far (surface distance) from the target should the bomb be released? The horizontal motion is independent of the vertical motion… except that the time that the projectile is in the air MUST be the same for both dimensions. The horizontal frame of reference is not accelerating therefore… d = v (t) d = 115m/s (14.6s) d = 1680 m 115 m/s 1050 m

  12. YOU TRY • PG. 102 #1-4

  13. What if the object is projected at an angle? • Then we can use the following equations: • Vx=Vi(cosΘ) • Δx=Vi(cos Θ) Δt • Vyf2=Vi2(sin Θ)2 - 2aΔy • Δy=Vi(sin Θ) Δt - 1/2a(Δt)2 • Vyf=Vi(sinΘ) - aΔt

  14. PG. 104 #1-5

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