projectile motion - angled launches fill in the component and resultant velocities of the thrown...

Projectile Motion - Angled Launches

Fill in the component and resultant velocities of the thrown projectile as it ascends and descends.

Assume that g = - 10 m/s2

30.4 m/s

20 m/s

10 m/s

5 m/s

20.6 m/s

11.2 m/s

5 m/s

11.2

20.6

30.4

Determine the maximum height reached by the projectile

vvi = 30 m/s vvtop = 0 m/s g = -10 m/s2 H = ?

vvf2 = vvi

2 + 2gH

H = (vvtop2 - vvi

2) / 2g

H = (0m/s2 - 30m/s2) / 2(-10m/s2) = 45m

Determine the time in the air and the range

vvf = vvi + gt So ….-vvi = vvi + gtAir

-2vvi = gtAIr So.. tAir = -2vvi / g = -2(30m/s)/ -10m/s2

But….vvf = - vvi

= 6s

dH = vH t

R = vH tAir = (5m/s) (6s) = 30m

Projectile Motion - Angled Launches

5m

20m

t = 1s

t = 2s

Projectile Motion - Angled Launches

Projectile Motion - Angled Launches

If there wasn’t any gravity the banana would follow a straight line trajectory and of course reach the monkey which would also not fall below its inertial position

If there is gravity the banana falls below its inertial position by 1/2gt2. If the monkey lets go at the same time the banana is fired, he too will fall below his rest position by 1/2gt2.

Projectile Motion - Angled Launches

If the keeper fires the banana at a slower speed it will still fall below its inertial position by 1/2gt2 but the horizontal component of its speed will be smaller so the banana will take longer to reach the monkey. If too slow the monkey will hit the ground before the banana can get across.

Projectile Motion - Angled Launches

Conceptual Example I Shot a Bullet into the Air...

Suppose you are driving a convertible with the top down.The car is moving to the right at constant velocity. You pointa rifle straight up into the air and fire it. In the absence of airresistance, where would the bullet land – behind you, aheadof you, or in the barrel of the rifle?

Bullet retains the horizontal velocity of the car so relative to a person in the car it will appear to go straight up and down.

Example The Height of a Kickoff

A placekicker kicks a football at and angle of 40.0 degrees andthe initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.

ov

oxv

oyv

14m soyv sinov 22m s sin 40

oxv cosov 22m s cos 40 17 m s

y ay vy voy t

? - 9.8 m/s2 0 14 m/s

y ay vy voy t? -9.80 m/s2 0 14 m/s

yavv yoyy 222 y

oyy

a

vvy

2

22

10 m

2

2

0 14m s

2 9.8m sy

The Time of Flight of a Kickoff

What is the time of flight between kickoff and landing?

y ay vy voy t0 -9.80 m/s2 14 m/s ?

y ay vy voy t0 -9.80 m/s2 14 m/s ?

221 tatvy yoy

2221 sm80.9sm140 tt

t2sm80.9sm1420

s 9.2t

The Range of a Kickoff

Calculate the range R of the projectile.

17 m s 2.9 s oxx v t 49 m

Conceptual Example Two Ways to Throw a Stone

From the top of a cliff, a person throws two stones. The stoneshave identical initial speeds, but stone 1 is thrown downwardat some angle above the horizontal and stone 2 is thrown atthe same angle below the horizontal. Neglecting air resistance,which stone, if either, strikes the water with greater velocity?

Both stones strike the water with the same velocity

Shoot the Monkey - Problem

A projectile is shot at the center of a target 2m away and 4m high. The target drops at the same time the projectile leaves the gun and is hit after it falls 1m. Determine the initial velocity of the projectile (speed and angle ).

H =4m

dv

dH = 2m

1m

vi

Shoot the Monkey - Solution

Horizontal Vertical

dH = 2m

dH = vH t = (vi cos) t

t = dH (vi cos)

dv = 3m g = -9.8 m/s2

dv = (vi sin) t + 1/2 g t2

dv = (vi sin) dH + 1/2 g ( dH )2

(vi cos) vi cos

= tan-1 (H /dH) = tan-1 (4m /2m) = 63.430

dv = vvi t + 1/2 g t2

Shoot the Monkey - Solution

dv = (sin) dH + 1/2 g ( dH )2

(cos) vi cos

3m = (tan 63.430) (2m) + 1/2 (- 9.8 m/s2) ( 2m )2

vi cos 63.430

3m = 3.999m + - 97.966 m3/s2 vi

2

- 0.999 m = -97.966 m3/s2 vi

2

vi = (- 97.966 m3/s2 ) -0.999m

= 9.9 m/s

Shoot the Monkey - Check

Horizontal Vertical

dH = 2m

dH = vH t

vH = dH t

dFF = 1m

dFF = 1/2 g t2

vH = 2m 0.45s

= 4.43 m/s

The projectile “falls” below its path (dashed) due to gravity. The freefall distance (dFF) is therefore 1m.

t = 2 (dFF) / g

t = 2(1m) / 9.8m/s2 = 0.45 s

Shoot the Monkey - Check

vi

vH

vvi

= tan-1 (H / dH) = tan-1 (4m / 2m)

= 63.430

Cos = vH / vi

vi = vH / Cos = 4.43m/s / Cos (63.430)

vi = 9.9 m/s