Zero Launch Angle. since θ=0, then v oy =0 and v ox = v o. The time required to reach the water. independent of v o!!

Transcription

1 Zero Launch Angle y h since θ=0, then v oy =0 and v ox = v o and based on our coordinate system we have x o =0, y o =h x The time required to reach the water independent of v o!! 1

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3 Combining Eliminating t and solving for y as a function of x: The landing point can be found by setting y = 0 and solving for x: x 3

4 Clicker You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all

5 You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will: a) quickly lag behind the plane while falling b) remain vertically under the plane while falling c) move ahead of the plane while falling d) not fall at all Both the plane and the package have the same horizontal velocity at the moment of release. They will maintain this velocity in the x-direction, so they stay aligned.

6 Airplane moving with a constant speed at some altitude. In the course of its flight, the plane drops a package from its luggage compartment. Motion Airplane and package before being dropped x= x o + v ox t Motion of package after being dropped x= x o + v ox t y= y o + v oy t -1/2 gt 2 Follow-up: what would happen if air resistance is present? 6

7 We return to our basic equations for general projectile motion Recall relationship: v 0x = v 0 cosθ and v 0y = v 0 sinθ This gives the equations of motion for the case x(t=0)=0, y(t=0) = 0: 7

8 Snapshots of a trajectory; For θ=35 o v o = 20m/s red dots are at times t = 0.5s, t = 1.0s, and t = 1.5s 0.5sec, 1.0sec, 1.5sec, 8

9 The total velocity at any point is found by vectorially adding the vertical component of the velocity, at that point, to the horizontal component of the velocity at that point. The horizontal velocity remains constant, because there is no acceleration in that direction. For a non zero launch angle, the initial vertical velocity is upward. However the vertical velocity gets smaller and smaller due to the downward acceleration of gravity V Vertical V total V Horizontal After reaching the maximum height, the projectile follows a zero launch angle trajectory. The vertical velocity is downward and the speed gets larger and larger. V Vertical V total V Horizontal 9

10 y component of velocity = 0 at the highest point Solve for t If t=0 during launch, this is time at the highest point Time is independent of the horizontal 10

11 Find the total elapse time T the projectile remains in the air If y=0 at launch time, we find the time when the projectile returns to the ground at y=0 Solving for t we find Total time is also independent of the horizontal component 11

12 Clicker Ignoring the air Punts resistance, which of the three punts has the longest time in the air (hang time)? a) b) c) d) all three have the same hang time h 12

13 Range: the horizontal distance a projectile travels If the initial and final elevation are the same: Assume the Projectile starts at the origin at t=0 y=0, solve for t Time when the projectile land (or returns to y=0) 13

14 Launch Angle (Deg) Launch Angle (Deg) Double solution function 0 to 90 degrees 14

15 Complementary angles will produce the same range. The maximum height will be different for the two angles. The times of the flight will be different for the two angles. Easy to see when the projectile initial angle is 45 the range is a maximum, however for range values less than the maximum there are 2 possible initial projectile angles and consequently 2 different trajectory paths. For example, 30 and 60 degrees projectiles find the same mark. The horizontal velocity component is greater for the 30 degree than the 60, however spends less time in the air. For 60 degrees, the trajectory spends more time in the air than the 30, but the horizontal velocity component is less than the