Jan 29: Projectile range and height problems

Components of the velocity vectorMaximum height and range for a

projectileExamples

Components of vectorsProjectile motion: rewrite in terms or horizontal and vertical components

€

v

€

vx = v cos(θ)

vy = v sin(θ)

v (without the vector) is the speed is the angle measured counterclockwise from the positive x direction

y

x

€

v = vx2 + vy

2

θ = arctanvyvx

⎛

⎝ ⎜

⎞

⎠ ⎟= tan−1 vy

vx

⎛

⎝ ⎜

⎞

⎠ ⎟

vx

vy

Basics of projectile motion

0

1

2

34

5

6

7

8

x

y

Tip: vy=0 at peak. Peak time is for t=vy0/g.Maximum height is

€

x = vx,0 t = v0 cos(θ0) t

"Airtime": t f = 2vy,0g

= 2v0 sin(θ0)

g

"Range"= (Horizontal velocity)• (Airtime) :

x f = v0 cos(θ0) • 2v0 sin(θ0)

g=vo

2 sin 2θ0( )g

vx0

vy0

€

v0

Maximum range for a given speed for 0=45°:

€

ymax − y0 =vy0

2

2g

€

xmax =v0

2

g

Projectile example

• Suppose a golf ball is struck at an angle 10° above the horizontal at a speed of 256 feet/second:* In the absence of air effects, how far away would it land?

*USGA standard test conditions: http://www.usga.org/equipment/protocols/alcod-PhaseII_rev1c.pdf

€

v0 = 256ft

s• 0.3048

m

ft= 78

m

s

Projectile example

• Suppose a golf ball is struck at an angle 10° above the horizontal at a speed of 256 feet/second:* In the absence of air effects, how far away would it land?

*USGA standard test conditions: http://www.usga.org/equipment/protocols/alcod-PhaseII_rev1c.pdf

€

Range =(78

m

s)2 sin(20°)

9.8m

s2

= 212 m€

"Range"= (Horizontal velocity)• (Airtime) :

x f = v0 cos(θ0) • 2v0 sin(θ0)

g=vo

2 sin 2θ0( )g

Interestingly, the acceptable range is up to 320 yard or 292 meter. The aerodynamic lift produced by the backspin of the ball (42 rev/s) makes it travel farther than it would in vacuum!

€

v0 = 256ft

s• 0.3048

m

ft= 78

m

s

Golfing on the moon

• Suppose the golf ball was struck on the Moon instead of the Earth. The acceleration due to the Moon’s gravity is approximately 1/6 that of Earth’s. Given the answer for Earth’s gravity, approximately how far would the ball travel on the Moon (which has effectively no atmosphere)?

Golfing on the moon

• Suppose the golf ball was struck on the Moon instead of the Earth. The acceleration due to the Moon’s gravity is approximately 1/6 that of Earth’s. Given the answer for Earth’s gravity, approximately how far would the ball travel on the Moon (which has effectively no atmosphere)?

Since range is

€

x f =vo

2 sin 2θ0( )g

For a value of g 1/6 as big, the range would be6 times farther (due to longer flight time).That would be over 1200 meter (3/4 mile)!

Vertical jump

From the NFL’s Combine* event description: “Players stand flat-footed in front of a pole that has plastic flags sticking out of it. Players jump from a standing position and try to swat as many of those flags as they can.”

The best jumpers can jump 1.0 meters or higher (40 or more inches). Estimate their launch speed.

*http://www.nfl.com/combine

Vertical jump

From the NFL’s Combine* event description: “Players stand flat-footed in front of a pole that has plastic flags sticking out of it. Players jump from a standing position and try to swat as many of those flags as they can.”

The best jumpers can jump 1.0 meters or higher (40 or more inches). Estimate their launch speed.

*http://www.nfl.com/combine€

ymax − y0 =vy0

2

2g

vy0 = 2g(ymax − y0)

For a 1 meter vertical jump, we need :

vy0 = v0 = 2 9.8m

s2

⎛

⎝ ⎜

⎞

⎠ ⎟(1m) = 4.4

m

s.

• In another Combine test, participants try to jump as far horizontally from a standing position with both feet together (this was an early modern Olympic event, and still competed in Norway). Assuming an ideal launch angle (45°), estimate the jump range given the same launch velocity as for the vertical jump.

• Assuming an ideal launch angle (45°), estimate the jump range given the same launch velocity as for the vertical jump (4.4 m/s).

Broad jump

€

xmax =v0

2

g=

4.4m

s

⎛

⎝ ⎜

⎞

⎠ ⎟2

9.8m

s2

= 2.0 m

Proficient broad jumpers land 11 feet away (about 3.6 meters). Can you think of reasons why actual jumps are longer than you just calculated? It may be useful to note that the peak height achieved in an 11 foot long jump is almost as high as during the vertical jump. Given that fact, what do you estimate the vertical and horizontal components of the velocity are for a good standing broad jump?

They must each be about 4 m/s.

Olympic long jump

• The long jump as currently competed allows a running start to a fixed take-off point. Assuming the same “airtime” as a 1.0 meter vertical jump, what horizontal velocity would be required to equal the current world record, 8.95 meter (Mike Powell, 1991*).

*Beating Carl Lewis for the World Championship, in what was probably the greatest head-to-head competition ever in a field event:http://news.bbc.co.uk/sport/hi/english/static/in_depth/athletics/2001/world_athletics/long_jump.stm €

t f =2vy0

g=

2 4.4m

s

⎛

⎝ ⎜

⎞

⎠ ⎟

9.8m

s2

= 0.90 s

x f = vx0t f = 8.95 m

vx0 =8.95 m

0.90 s= 9.9

m

s

seems reasonable (somewhat

slower than the best sprinters)