Loading...

PPT – Projectile Motion Examples PowerPoint presentation | free to download - id: 3f422f-OGQ3N

The Adobe Flash plugin is needed to view this content

Projectile Motion Examples

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

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

Conceptual Example 3-8

- Demonstration!!

v0x ?

Conceptual Ex. 3-9 Wrong Strategy

- Shooting the Monkey!!
- Video Clip!!

Example 3-10

- Range (R) of projectile ? Maximum horizontal

distance before returning to ground. Derive a

formula for R.

- 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)

Example 3-11, A punt!

- v0 20 m/s, ?0 37º
- vx0 v0cos(?0) 16 m/s, vy0 v0sin(?0) 12 m/s

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!!

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?

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?

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?

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?

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

Problem 46

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.

- 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.

- 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.