Projectile Motion. break the initial velocity into its 2 components, horizontal and vertical

Transcription

1 Projectile Motion when an object that moves through space is acted upon by Earth's gravity Ex. A football player kicks a football through the end zone for a field goal Of course there is an initial velocity, but then only gravity... Now we have 2 dimensions and we break the initial velocity into its 2 components, horizontal and vertical

2 IMPORTANT ; g only acts vertically, so the vertical velocity will change but the horizontal velocity remains constant. if the initial velocity is 20 m/s at an angle of 15 degrees what are the horizontal and vertical components? horizontal componenet 20 cos 15 = 19 m/s vertical component 20 sin 15 = 5.2 m/s Once we have the components: we can use these values in the same equations we used for rectilinear motion to find variables such as time, distance and velocity. Keep in mind the final velocity may also have 2 components and you may need to find the resultant vector.

3 Let's start with an example that is launched horizontally. That means, there is no initial velocity for the y component, only the horizontal component. A marble rolls of the table with a horizontal velocity of 1.2m/s It lands in a cup 0.51m from the table's edge. How high is the table? Horizontal vi = 1.2 m/s vf = 1.2 m/s di = 0m df = 0.51m knowing the velocity is constant I know that v = d/t so I can find t t = d/v = 0.51m/ (1.2m/s) = 0.43s Now I have time Vertical vi = 0 m/s g = -9.8 m/s 2 di = x df = 0m t = 0.43s What forumal? df = di + vit + 1/2at 2 0m = x + 0*t -4.9t 2 x = 4.9*(.43) 2 x = 0.91m The table is 0.91 m high Helpful Hints: It is common to put the origin of the cartesian plane at ground level where x = 0 is at the launch site Projectile objects leaving the ground will reach the peak height at half the total time If taking off from a different level (height), the time it takes to return to it's original height is twice the time it takes to reach the peak height Emmanuel Zacchini, the famous human cannonball, was fired out of a cannon with a speed of 24.0 m/s at an angle of 40.0 degrees to the horizontal. If he landed in a net 56.6 m away at the same height from which he was fired, how long was Zacchini in the air? Find the horizontal speed, and given the distance find t. Vh = 24.0 m/s cos 40 = 18.4 m/s t = d/v (velocity is constant) = 56.6m/ 18.4 m/s =3.08s

4 Evel Knievel, successfully jumped 69.5 m over a Grand Canyon gorge. Assuming that he started and landed at the same level and was airborne for 3.66s, what height from his starting point did this daredevil achieve? Knowing the max height was at the halfway point, we know that t = 1.83s g= -9.8m/s 2 What equation df = di + vit + 1/2at 2 If we say vi = 0 at the halfway point then di= x and df = 0 Then 0 = x + 1/2(-9.8m/s 2 )(1.83s) 2 x = 16.4m so the motorcycle reached a height of 16.4m Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly for lunch. If the lily pads are spaced 2.4 m apart, and Ferdinand jumps with a speed of 5.0m/s, taking 0.60 s to go from lily pad to lily pad, at what angle must Ferdinand make each of his jumps? df= 2.4 di = 0m vi = 5.0m/s t = 0.60s horizontal v =d/t 2.4/0.69 = 4 m/s The horizontal component of vi = vicosθ = vx 5.0 m/s * cos θ = 4 m/s cos θ = 4/5 θ = cos -1 (4/5) = 37 degrees to horizontal We often find the time first and then plug the time in to find other variables. Let's look at the examples in the textbook p A soccer player kicks the ball with a velocity of 20.0 m/s at a 25.0 degree angle above the horizontal How long does it take for the ball to reach its maximum height?

5 A soccer player kicks the ball with a velocity of 20.0 m/s at a 25.0 degree angle above the horizontal How long does it take for the ball to reach its maximum height? vix = 20.0 cos 25.0 v is constant viy = 20.0 sin 25.0 v2 = 0m/x What formula? #1 Vf = Vi + at 0m/s = 8.45m/s - 9.8t -8.45/-9.8 = t t = 0.862s What is the balls flight time? We know it should be 2 * 0.86 seconds or 1.72 seconds We can also use the formula yf = yi + vit +1/2at 2 Yf will be 0 again (it is back on the ground) 0 = m/s t m/s 2 t 2 0 = t( t) t = 0 (when the ball was kicked 9.45/4.9 = t t= 1.72s What distance does the ball travel horizontally? We know the time is 1.72s The horizontal velocity is 20.0m/s cos 25.0 = 18.1m/s (and its constant so vi and vf = 18.1m/s) v = d/t and d = vt so d = 18.1m/s*1/72s = 31.2m

6 How do we get the resultand vector and the angle? 8.45m/s θ 18.1m/s Pythagorean theorem: find the hypotenuse = 20.0m/s tan θ = -8.45/181 θ = = from the horizontal. Example B: Here is the question: Try it and check your answer in the text book. From a roof 50.0m high, a ball is thrown with a velocity of 5.00m/s at 25.0 o above the horizontal. List all your knowns for vertical and horizontal: a) how long does it take for the ball to reach the ground? ( hint: you will need to use the quadratic equation: use the positive value of t) b) what is the magnitude of the ball's velocity at the moment it touches the ground? What is the direction of the velocity vector at this moment c) what is the maximum height reached by the ball? Next time; Problems: section 11.2, 251 all if we changed only the angle of the launch, what would happen to the maximum height? and the horizontal distance? design your own lab...

7 d) what is the value of the ball's velocity at the moment it touches the ground? give the magnitude and direction... (yes it is a vector with an x component and a y component) we know vxf = 18.1 m/s vyf =? we have a and t and vi so we can use: vf = vi + at = 8.45m/s + (-9.8m/s 2 )(1.72s) = -8.45m/s (directed down) 8.45m/s 18.1m/s How do we get the resultand vector and the angle?