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Published byRebecca Norris Modified over 9 years ago
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Projectile Motion From the ground From a cliff
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Kicked off a cliff Review: Objects Kicked off a cliff have: X-dirY-dir v= const.vi=0 m/s a= 0 m/s 2 a = -9.8 m/s Δx= positiveΔy= negative v f = negative
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Kicked off a cliff This is a combination of basic Horizontal Motion & Vertical Motion
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Projectile Motion- Level Ground Examples: Footballs, golf balls, bullets, catapults. There is no displacement in the Δy direction between the beginning “sea level” and the ending “sea level”
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Projectile Motion- Level Ground This is a combination of Vertical Motion & Horizontal Motion
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Projectile Motion- Level Ground What do we know about: X-directionY-direction
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Projectile Motion- Level Ground What do we know about: X-directionY-direction Δx right = +Δy= 0 v=constantv i = + a= 0m/s 2 v f = - a= -9.8m/s 2
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Projectile Motion- Level Ground Attacking the problem: Initial velocities will be given in vectors: 25 m/s @ an angle of 30˚. 30˚
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Projectile Motion- Level Ground Resolve the vector into components to determine Initial Velocity in the X and Y directions. 30˚ 25 m/s
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Projectile Motion- Level Ground Resolve the vector into components to determine Initial Velocity in the X and Y directions. X-dir Y-dir 30˚ 25 m/s vxvx vyvy
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Projectile Motion- Level Ground What do we know about: X-directionY-direction Δx right = +v x Δy= 0 v=constantv i = +v y a= 0m/s 2 v f = -v y a= -9.8m/s 2 Now you are ready to solve the problem…
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