Physics 13 General Physics 1 Kinematics of Motion Projectile Motion MARLON FLORES SACEDON

Motion in Two or Three Dimensions: Projectile Motion Introduction A body that moves through space usually has a curved path rather than a perfectly straight one. Projectile anybody that is given an initial velocity and then allowed to move under the influence of gravity. The path followed by a projectile is called its trajectory. x x x

Motion in Two or Three Dimensions: Projectile Motion Assumptions Air resistance is neglected (𝐴𝑖𝑟 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛=0) Component of acceleration along x axis is zero ( 𝑎 𝑥 =0). So, motion is uniform. Component of acceleration along y axis is gravitational acceleration ( 𝑎 𝑦 =−𝑔). So, Free-fall. x y 5. 𝑣= 𝑣 𝑥 2 + 𝑣 𝑦 2 𝑣 𝑦 =0 𝑣= 𝑣 𝑥 𝑣 𝑥 𝑣 𝑦 𝑣 6. 𝑥= 𝑣 𝑜𝑥 𝑡+ 1 2 𝑎 𝑥 𝑡 2 𝑣 𝑥 𝑣 𝑦 𝑣 𝑣 𝑥 −𝑣 𝑦 𝑣 = 𝑣 𝑜 𝑐𝑜𝑠𝜃.𝑡 x y 7. t= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 𝑣 𝑜 Eliminate 𝑡 in Eq.7 in Eq.8 to come-up with time-independent equation 𝑣 𝑜𝑥 𝑣 𝑜𝑦 𝜃 8. 𝑦= 𝑣 𝑜𝑦 𝑡+ 1 2 𝑎 𝑦 𝑡 2 = 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 1. 𝑣 𝑜𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 3. 𝑣 𝑥 = 𝑣 𝑜𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 Time-independent Equation of Projectile 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 2. 𝑣 𝑜𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃 4. 𝑣 𝑦 = 𝑣 𝑜𝑦 −𝑔𝑡 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡

Motion in Two or Three Dimensions: Projectile Motion Formulas for Projectile Motion 1. 𝑣 𝑜𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 x y 𝑣 𝑜 𝜃 𝑣 𝑜𝑥 𝑣 𝑜𝑦 𝑣 𝑥 −𝑣 𝑦 𝑣 2. 𝑣 𝑜𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃 3. 𝑣 𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 4. 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 5. 𝑣= 𝑣 𝑥 2 + 𝑣 𝑦 2 6. 𝑥= 𝑣 𝑜 𝑐𝑜𝑠𝜃.𝑡 7. 𝑡= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 8. 𝑦= 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 The Time-independent Equation of Projectile 9. 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃

Motion in Two or Three Dimensions: Projectile Motion Maximum Range, Time to reach the Range Time to peak, & Maximum height B. For Time to reach the Range ( 𝑇 𝑅𝑎𝑛𝑔𝑒 ) From Formula # 6, Therefore: x y Maximum height ( 𝑦 𝑚𝑎𝑥 ) = ? 𝑥=R ( 𝑇 𝑅𝑎𝑛𝑔𝑒 )= 𝑅 𝑣 𝑜 𝑐𝑜𝑠𝜃 𝑣 𝑜 = 𝑣 𝑜 𝒚 𝒎𝒂𝒙 Time to peak ( 𝑇 𝑝𝑒𝑎𝑘 ) = ? 𝑡= 𝑇 𝑅𝑎𝑛𝑔𝑒 𝜃=𝜃 C. For Time to Peak ( 𝑻 𝒑𝒆𝒂𝒌 ): Velocity along y is zero, Formula #4 becomes 𝑣 𝑜 𝑣 𝑜𝑥 𝑣 𝑜𝑦 Time to reach the Range ( 𝑇 𝑅𝑎𝑛𝑔𝑒 ) = ? 𝜃 0= 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔 𝑻 𝒑𝒆𝒂𝒌 𝑻 𝒑𝒆𝒂𝒌 = 𝑣 𝑜 𝑠𝑖𝑛𝜃 𝑔 Range 𝑅 =? So: 0=R𝑇𝑎𝑛𝜃− 𝑔 𝑅 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 Formula needed 4. 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 A. For Range: 𝑔 𝑅 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 =R𝑇𝑎𝑛𝜃 D. For Maximum Height 𝒚 𝒎𝒂𝒙 : 8. 𝑦= 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 From Formula # 9, 𝑥=𝑅 From Formula # 8, substitute 𝑻 𝒑𝒆𝒂𝒌 to 𝑡 7. 𝑡= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 𝜃=𝜃 𝑅= 2 𝑣 𝑜 2 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃 𝑔 𝒚 𝒎𝒂𝒙 = 𝑣 𝑜 𝑠𝑖𝑛𝜃. 𝑣 𝑜 𝑠𝑖𝑛𝜃 𝑔 − 1 2 𝑔 ( 𝑣 𝑜 𝑠𝑖𝑛𝜃 𝑔 ) 2 The Time-independent Equation of Projectile 9. 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 𝑣 𝑜 = 𝑣 𝑜 𝑦=0 𝑅= 𝑣 𝑜 2 𝑠𝑖𝑛2𝜃 𝑔 𝒚 𝒎𝒂𝒙 = 𝑣 𝑜 2 𝑠𝑖𝑛 2 𝜃 2𝑔

Motion in Two or Three Dimensions: Projectile Motion Problem Exercise A tennis ball rolls off the edge of a table top 0.75 m above the floor and strikes the floor at a point 1.40 m horizontally from the edge of the table. Ignore air resistance. a). Find the magnitude of the initial velocity. b). Find the time of flight c). Find the magnitude and direction of the velocity of the ball just before it strikes the floor. a). The magnitude of the initial velocity ( 𝑣 𝑜 ) From: The Time-independent Equation of Projectile 9. 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 𝑥=1.40𝑚 𝑦=−0.75𝑚 𝑣 𝑜 𝑣 𝑜 =? 𝜃=0 −0.75=1.40𝑇𝑎𝑛0− 9.81( 1.40 2 ) 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 0 𝑣 𝑜 = 9.81( 1.40 2 ) 2(0.75) =3.58 m/s From Formula 7. 𝑡= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 b). Find the time of flight 𝑡 𝑓𝑙𝑖𝑔ℎ𝑡 = 1.40 3.58𝑐𝑜𝑠0 =0.39 𝑠𝑒𝑐 0.75𝑚 c). Find the magnitude and direction of the velocity of the ball just before it strikes the floor. 𝑡 𝑓𝑙𝑖𝑔ℎ𝑡 3. 𝑣 𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 4. 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 𝑣 𝑥 = 𝑣 𝑜 =3.58𝑚/𝑠 1.40𝑚 𝛽 𝑣 𝑓 𝑣 𝑦 =3.58 𝑠𝑖𝑛0 −9.81(0.39)=-3.83m/s 𝑣 𝑓 = 3.58 2 + 3.83 2 =5.24𝑚/𝑠 𝛽= 𝑇𝑎𝑛 −1 3.83 3.58 =46.93𝑜

Motion in Two or Three Dimensions: Projectile Motion

Motion in Two or Three Dimensions: Projectile Motion 𝑣 𝑜 = ?

Motion in Two or Three Dimensions: Projectile Motion 𝑣 𝑜 = ? Height cliff ℎ= ?

Motion in Two or Three Dimensions: Projectile Motion Will the snowball hit the man? 1.90 m

Motion in Two or Three Dimensions: Projectile Motion What distance D from the dock should the ship be when the equipment is thrown? Ignore air resistance.

Motion in Two or Three Dimensions: Projectile Motion river 53o 15m 100m 40m 𝑣 𝑜 =?

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