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1. Projectile Motion at an Angle Katya Fatakhova Lisa Levenson Kaitlin Magee Daniel Wittenberg

2. The Basics • Figuring out the different parts of a projectile launched at an angle require a few more steps than when it is just launched horizontally • There are more concepts to it and there are different times that are used for different equations.  • The best example would be to look at how a ball is kicked

3. -When the ball is first hit, it is at the fastest speed it can go, this is called theinstantaneous velocity. -Once it leaves your foot, gravity acts on the vertical velocity of the ball. This number will decrease as the ball gets higher  because gravity continues to pull down at it. Then it will increase to its original value as the ball gets closer to the ground. - When the ball is at its highest peak, the vertical velocity is ZERO! This is because it is neither moving up nor down.  - The horizontal component of velocitystays the same all throughout.

4. Figuring out the components -When you have the instantaneous velocity, to find out what the vertical component of velocity is, you use the equation: vy=vi(sinθ) -When you have the instantaneous velocity, to find out what the horizontal component of velocity is, you use the equation: vx=vi(cosθ) ﻿vy v v vx

5. -Since the ball makes an arc shaped movement after it is kicked, it also has a maximum height as well as a range, which is the total distance that it traveled.

6. Figuring out the components Part II -When finding the maximum height (dy or hmax),the equation used is: dy=vyt+(1/2)ayt2 vy is the vertical component of velocity and the ay is the gravity that pushes down on the vy, which is (-9.81 m/s2) As for the t (time) used in this equation, it is not the same time as used in horizontal projectile motion (cont. on next slide)

7. All About Tup -Since the time it takes half to the time reach the maximum height as it does to complete a projection, the term tup(time up) is used. The equation for this t is: t=(-vy)/(a) Because of the parabolic curve of the motion, the ball is at its highest point halfway through the complete cycle. Therefore, tup is the time used to find dmax dy-max = (1/2)(a)(tup)2 (When the object is at its highest point, its vertical velocity is zero.)

8. Range -Now to find out the range, the equation used is: dx=vxt -The vx is the horizontal component of velocity and the t this time is the total time, so you could use either the equation  ttotal = √(2d/a) or you can just use the tup from before and multiply it by 2 (tup = ttotal/2)

9. *NOTE* The maximum range of a projectile launched at a given speed is achieved when the angle with the horizontal is 45° (cos 45º) = 1    Animation:  NEXT SLIDE

10. 183. The maximum height reached by the ball is approximately (1) 1.2 m (2) 2.5 m (3) 4.9 m (4) 8.5 m

11. 183. The maximum height reached by the ball is approximately (1) 1.2 m (2) 2.5 m (3) 4.9 m (4) 8.5 m

12. Explanation -So you have to use the equation: dy=vyt+(1/2)ayt2 -You plug in all of the values you know:  (the tup is the t given divided by two so it's 1/2) dy=(4.9 m/s)(0.5 s)+(1/2)(-9.81 m/s2)(0.5 s)2 dy= 1.2 m

13. Projectile Motion at an Angle An  archer uses a bow to fire two similar arrows with the same string force. One arrow is fired at an angle of 60 degrees with the horizontal, and the other is fired at an angle of 45 degrees with the horizontal. Compared to the arrow fired at 60 degrees, the arrow fired at 45 degrees has a  (A) Longer flight time and shorter horizontal range (B) Shorter flight time and longer horizontal range (C) Shorter flight time and shorter horizontal range (D) Longer flight time and longer horizontal range

14. Projectile Motion at an Angle An  archer uses a bow to fire two similar arrows with the same string force. One arrow is fired at an angle of 60 degrees with the horizontal, and the other is fired at an angle of 45 degrees with the horizontal. Compared to the arrow fired at 60 degrees, the arrow fired at 45 degrees has a  (A) Longer flight time and shorter horizontal range (B) Shorter flight time and longer horizontal range (C) Shorter flight time and shorter horizontal range (D) Longer flight time and longer horizontal range

15. Explanation dx = (vx)(time)+(1/2)at2 - There is no horizontal acceleration in this problem, so the horizontal velocity will stay constant(neglecting air resistance) So the equation becomes, dx = vx(t) - To find vx, you have to multiply the original diagonal velocity by cosθ cos(45º) = 1- This means that a projectile fired at 45º will always have the longest horizontal distance possible. cos (60º) = 0.5 - If the two arrows were fired with the same initial velocity, the one fired at 45º has a larger horizontal velocity, and therefore travels farther horizontally.

16. Explanation (continued) From the formula d=vi(t) + (1/2)at2 we find that  t = √(2dy/a) . Firing the arrow at 60º causes dy to be increased, and therefore makes the flight time longer. Summary - The larger the angle, the longer the flight time.  - The closer the angle is to 45º, the longer the horizontal distance

17. Review - Short Answer An airplane flying horizontally at 150m/s drops a bomb from a height of 2500m. How far in front of the target must the bomb be released? This problem must be solved in two steps: 1) Find the time it takes for the bomb to hit the ground dy = viy + (1/2)at2 (The initial y velocity is 0.) 2,500m = (1/2)(9.81m/s2)t2 2,500m = 4.9m/s2(t)2 2,500m/4.9m/s2 = t2 t =√(510s2) t = 23s

18. THE END