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Projectile Motion Examples

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Projectile Motion Examples Example 3-6: Driving off a cliff!! y is positive upward, y0 = 0 at top. Also vy0 = 0 Example 3-7: Kicked football A ... – PowerPoint PPT presentation

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Title: Projectile Motion Examples


1
Projectile Motion Examples
2
Example 3-6 Driving off a cliff!!
A movie stunt driver on a motorcycle speeds
horizontally off a 50.0-m-high cliff. How fast
must the motorcycle leave the cliff top to land
on level ground below, 90.0 m from the base of
the cliff where the cameras are?
  • y is positive upward, y0 0 at top. Also vy0
    0

vx vx0 ? vy -gt x vx0t, y - (½)gt2 Time
to Bottom t v2y/(-g) 3.19 s vx0 (x/t)
28.2 m/s
3
Example 3-7 Kicked football
  • A football is kicked at an angle ?0 37.0 with
    a velocity of 20.0 m/s, as shown. Calculate
  • a. Max height. b. Time when hits ground. c.
    Total distance traveled in the x direction. d.
    Velocity at top. e. Acceleration at top.
  • ?0 37º, v0 20 m/s
  • ? vx0 v0cos(?0) 16 m/s, vy0 v0sin(?0) 12 m/s

4
Conceptual Example 3-8
  • Demonstration!!

v0x ?
5
Conceptual Ex. 3-9 Wrong Strategy
  • Shooting the Monkey!!
  • Video Clip!!

6
Example 3-10
  • Range (R) of projectile ? Maximum horizontal
    distance before returning to ground. Derive a
    formula for R.

7
  • Range R ? the x where y 0!
  • Use vx vx0 , x vx0 t , vy vy0 -
    gt
  • y vy0 t (½)g t2, (vy) 2
    (vy0)2 - 2gy
  • First, find the time t when y 0
  • 0 vy0 t - (½)g t2
  • ? t 0 (of course!) and t (2vy0)/g
  • Put this t in the x formula x vx0 (2vy0)/g ?
    R
  • R 2(vx0vy0)/g, vx0 v0cos(?0), vy0 v0sin(?0)
  • R (v0)2 2 sin(?0)cos(?0)/g
  • R (v0)2 sin(2?0)/g (by a trig identity)

8
Example 3-11, A punt!
  • v0 20 m/s, ?0 37º
  • vx0 v0cos(?0) 16 m/s, vy0 v0sin(?0) 12 m/s

9
Proof that projectile path is a parabola
  • x vx0 t , y vy0 t (½)g t2
  • Note The same time t enters both equations!
  • ? Eliminate t to get y as a function of x.
  • Solve x equation for t t x/vx0
  • Get y vy0 (x/vx0) (½)g (x/vx0)2
  • Or y (vy0 /vx0)x - (½)g/(vx0)2x2
  • Of the form y Ax Bx2
  • A parabola in the x-y plane!!

10
Ex. 3-12 a) Rescue Helicopter Drops Supplies
A rescue helicopter wants to drop a package of
supplies to isolated mountain climbers on a rocky
ridge 200 m below. If the helicopter is traveling
horizontally with a speed of 70 m/s (250 km/h),
a) How far in advance of the recipients
(horizontal distance) must the package be
dropped?
11
Ex. 3-12 b), c) Rescue Helicopter Throws Supplies
A rescue helicopter wants to get a package of
supplies to isolated mountain climbers on a rocky
ridge 200 m below. The helicopter is traveling
horizontally with a speed of 70 m/s (250 km/h),
b) Someone in the helicopter throws the
package a horizontal distance of 400 m in advance
of the mountain climbers. What vertical velocity
should the package be given (up or down) so that
it arrives precisely at the climbers position?
c) With what speed does the package land?


12
Thats Quite an Arm!
Problem A stone is thrown from the top of a
building at an angle ?0 26 to the horizontal
and with an initial speed v0 17.9 m/s, as in
the figure. The height of the building is 45.0
m. a) How long is the stone "in
flight"? b) What is the speed of the
stone just before it strikes the
ground?
13
Example The Long Jump
Problem A long-jumper leaves the ground at angle
?0 20 above the horizontal at a speed of v0
8.0 m/s. a. How far does he jump in the
horizontal direction? B. What is his maximum
height?
14
Stranded Explorers
Problem An Alaskan rescue plane drops a package
of emergency rations to a stranded party of
explorers, as shown in the picture. If the plane
is traveling horizontally at v0 42.0 m/s at a
height h 106 m above the ground, where does the
package strike the ground relative to the point
at which it is released?
v0 42 m/s
h
15
Problem 46
16
Chapter 3, Problem 46 Solution
  • Choose the origin at ground level, under where
    the projectile is launched, up to be the
    positive y direction. For the projectile
  • a. The time to reach the ground is found from Eq.
    2-12b, with final height 0. Choose positive
    time since the projectile was launched at time t
    0.
  • b. The horizontal range is found from the
    horizontal motion at constant velocity.

17
  • c. At the instant just before the particle
    reaches the ground, the horizontal component of
    its velocity is the constant
  • The vertical component of velocity is found from
    Eq. 2-12a
  • d. The magnitude of the velocity is found from
    the x and y components calculated in part c.
    above.

18
  • e. The direction of the velocity is
  • so the object is moving
  • f. The maximum height above the cliff top reached
    by the projectile will occur when the y-velocity
    is 0, and is found from Eq. 2-12c.
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