# Projectile Motion - Angled Launches Fill in the component and resultant velocities of the thrown projectile as it ascends and descends. Assume that g =

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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/s 2 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 v vi = 30 m/s v vtop = 0 m/s g = -10 m/s 2 H = ? v vf 2 = v vi 2 + 2gH H = (v vtop 2 - v vi 2 ) / 2g H = (0m/s 2 - 30m/s 2 ) / 2(-10m/s 2 )= 45m Determine the time in the air and the range v vf = v vi + g  tSo ….-v vi = v vi + g  t Air -2v vi = g  t AIr So..  t Air = -2v vi / g = -2(30m/s)/ -10m/s 2 But….v vf = - v vi = 6s  d H = v H  t R = v H  t Air = (5m/s) (6s)= 30m

Projectile Motion - Angled Launches 5m 20m t = 1s t = 2s

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 1/2g  t 2. If the monkey lets go at the same time the banana is fired, he too will fall below his rest position by 1/2g  t 2.

Projectile Motion - Angled Launches If the keeper fires the banana at a slower speed it will still fall below its inertial position by 1/2g  t 2 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 point a rifle straight up into the air and fire it. In the absence of air resistance, where would the bullet land – behind you, ahead of 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 and the initial speed of the ball is 22 m/s. Ignoring air resistance, determine the maximum height that the ball attains.

yayay vyvy v oy t ?- 9.8 m/s 2 014 m/s

yayay vyvy v oy t ?-9.80 m/s 2 014 m/s

The Time of Flight of a Kickoff What is the time of flight between kickoff and landing?

yayay vyvy v oy t 0-9.80 m/s 2 14 m/s?

yayay vyvy v oy t 0-9.80 m/s 2 14 m/s?

The Range of a Kickoff Calculate the range R of the projectile.

Conceptual Example Two Ways to Throw a Stone From the top of a cliff, a person throws two stones. The stones have identical initial speeds, but stone 1 is thrown downward at some angle above the horizontal and stone 2 is thrown at the 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dv  d H = 2m 1m  vivi

Shoot the Monkey - Solution HorizontalVertical  d H = 2m  d H = v H  t = (v i cos  )  t  t =  d H (v i cos  )  d v = 3m g = -9.8 m/s 2  d v = (v i sin  )  t + 1/2 g  t 2  d v = (v i sin  )  d H + 1/2 g (  d H ) 2 (v i cos  ) v i cos   = tan -1 (H /  d H ) = tan -1 (4m /2m) = 63.43 0  d v = v vi  t + 1/2 g  t 2

Shoot the Monkey - Solution  d v = (sin  )  d H + 1/2 g (  d H ) 2 (cos  ) v i cos  3m = (tan 63.43 0 ) (2m) + 1/2 (- 9.8 m/s 2 ) ( 2m ) 2 v i cos 63.43 0 3m = 3.999m + - 97.966 m 3 /s 2 v i 2 - 0.999 m = -97.966 m 3 /s 2 v i 2 v i =  (- 97.966 m 3 /s 2 ) -0.999m = 9.9 m/s

Shoot the Monkey - Check HorizontalVertical  d H = 2m  d H = v H  t v H =  d H  t  d FF = 1m  d FF = 1/2 g  t 2 v H = 2m 0.45s = 4.43 m/s The projectile “falls” below its path (dashed) due to gravity. The freefall distance (  d FF ) is therefore 1m.  t =  2 (  d FF ) / g  t =  2(1m) / 9.8m/s 2 = 0.45 s

Shoot the Monkey - Check vivi  vHvH v vi  = tan -1 (H /  d H ) = tan -1 (4m / 2m)  = 63.43 0 Cos  = v H / v i v i = v H / Cos  = 4.43m/s / Cos (63.43 0 ) v i = 9.9 m/s

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