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19 Pages

### chapter3_part2.sxi

Course: PHY 101, Fall 2009
School: SUNY Buffalo
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Word Count: 845

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Motion 2D Projectile motion of an object. There is no acceleration in the x-direction. The acceleration in the y-direction is due to gravity. Concepts Object leaves the ground with some initial velocity and angle. The velocity in the xdirection is constant (since ax = 0). The velocity in the ydirection is changing with time (since ax is g). Vy is 0 at the highest point the object reaches. How to Approach...

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Motion 2D Projectile motion of an object. There is no acceleration in the x-direction. The acceleration in the y-direction is due to gravity. Concepts Object leaves the ground with some initial velocity and angle. The velocity in the xdirection is constant (since ax = 0). The velocity in the ydirection is changing with time (since ax is g). Vy is 0 at the highest point the object reaches. How to Approach These Problems As always, draw a diagram and label all the quantities given. Decide which direction is positive for both x and y directions. Break your vector describing the object's motion into its components. Solve for the x and y parts separately. Use the fact time is the same for each part to recombine them. Check you have answered what was asked and the units make sense and are correct. Baseball Thrown in the Air y ymax A baseball is thrown straight up in the air with an initial velocity of 10 m/s. Assuming no air resistance, how high above the ground will the ball go? Baseball Thrown in the Air (2) Given: Voy = 10 m/s Direction: straight up Acceleration: -9.8 m/s2 time distance in the ydirection y ymax Unknowns: Baseball Thrown in the Air (3) Look at your kinematic equations of motion: vy = v0y + ay t y = (v0y + vy)t y = v0y t + ay t2 vy2 = v0y2 + 2 ay y y = (vy2 - v0y2) / 2ay Since you have vy, v0y, and ay, you can solve the problem directly. Baseball Thrown in the Air (4) Solving for y we find: y = (02 - 102) m2/s2 / 2(-9.80) m/s2 y = -100 / - 19.6 m y = 5.10 m y ymax Golf Ball Hit with Driver A golf ball is hit using a number 1 driver. The ball leaves the grass at an angle of 50 with an initial velocity of 60 m/s. How far away will the ball land? y x Golf Ball Hit with Driver (2) Given: v0 is 60 m/s = 50 ay = g = -9.80 m/s2 ax = 0 m/s2 y x t y x Unknowns: Golf Ball Hit with Driver (3) Draw your diagram. Choose which directions are positive. In this case, I choose up and right to be positive. Break your initial vector into components x: 60.0 m/s cos 50 = 38.6 m/s y: 60.0 m/s sin 50 = 46.0 m/s Work separately with the x and y parts Golf Ball Hit with Driver (4) We are asked to find how far away the ball lands. This implies we need to solve for x. We will use what is commonly to referred as the Range Equation. x = v0x t Problem! We do not have the total time the ball is in the air! Look to the y-part for this information. Golf Ball Hit with Driver (5) We know the ball has an initial velocity in the ydirection v0x = 46.0 m/s. We also know the ball has a final velocity in the y-direction of 0 m/s when the ball reaches it's maximum height. When there is no air resistance and the point where the ball begins and ends in the y-direction is the same height, you can solve for half the problem (in this case the half where the ball is traveling upwards) and multiply the time by 2. Golf Ball Hit with Driver (6) Look at your kinematic equations of motion: vy = v0y + ay t y = (v0y + vy)t y = v0y t + ay t2 vy2 = v0y2 + 2 ay y t = (vy - v0y) / ay Since you have vy, v0y, and ay you can simply solve for the time it takes the ball to reach it's maximum height. This time, multiplied by 2, will give you the total time the ball is in flight. Golf Ball Hit with Driver (7) t = (vy - v0y) / ay t = (0 - 46.0) m/s / -9.80 m/s2 t = (-46.0) / -9.80 s t = 4.69 s for the time of upward flight total time t is t 2 = 9.38 s. Golf Ball Hit with Driver (8) We can now use this time, along with the initial velocity in the x-direction, to determine how far the ball travels in the x-direction. x = v0x t x = 38.6 m/s 9.38 s x = 362 m/s ...

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