SPH3U UNIVERSITY PHYSICS

Transcription

1 SPH3U UNIVERSITY PHYSICS KINEMATICS L (P.76-81) Projectile & The motion experienced by a dirt bike jumper is identical to that of a ball thrown up in the air at an angle. Both travel through a twodimensional curved path called a parabola. Any object that moves in response to gravity along a twodimensional curved trajectory is called a projectile. The motion of a projectile under gravity is called projectile motion. November 24, U1-1 Projectile & PROJECTILE object that moves along a twodimensional (2D) curved trajectory (a parabola) in response to gravity PROJECTILE MOTION the motion of a projectile under gravity November 24, U1-2 1

2 1. Imagine two rubber balls one is thrown horizontally while at the same time the other is allowed to drop to the ground. Which ball will hit the ground first? Both rubber balls will hit the ground at exactly the same time (if air resistance is negligible). November 24, U1-3 This is because both rubber balls experience a vertical acceleration due to gravity. Although the one rubber ball is projected horizontally, its horizontal motion does not affect its vertical motion (they are independent of each other). As a result, both rubber balls reach the ground at the same time. NOTE! The horizontal distance travelled by a projectile is known as the range and the amount of time the projectile is in the air is known as itstimeofflight. November 24, U1-4 TIMEOFFLIGHT(t flight ) the amount of time it takes for a projectile to complete its motion RANGE(R) the horizontal distance travelled by a projectile November 24, U1-5 2

3 Two-Dimensional Motion This situation is similar to a boat crossing a river. Consider a boat travelling at 12 km/h[n] across a 0.30 km wide river that is at rest. In this case it will take the boat h to cross the river (t=d/v). However, this is not realistic because most rivers are not at rest. Lets consider the same situation but with the river flowing at 24 km/h[e]. November 24, U1-6 Two-Dimensional Motion In this case, even though the boat is pointed due north, the boat does not arrive at it s home. Instead it lands some distance farther downstream. However, the time required to cross the river is still the same h because the boat s velocity and distance travelled in the northward direction have not changed. November 24, U1-7 Two-Dimensional Motion This is because the river s current is perpendicular to the boat s velocity it does not increase or decrease the boat s velocity in the northward direction. In other words, the velocity of the boat and the river are independent of one another. November 24, U1-8 3

4 2. Now consider an object being thrown horizontally out a window. There is one key difference between the river crossing scenario and this projectile motion scenario. What is it? With the river crossing scenario both velocities are constant. In this projectile motion scenario, while the horizontal velocity is constant, the vertical velocity changes because of the acceleration due to gravity. November 24, U1-9 As such, projectile motion problems are two-dimensional vector problems. Î Horizontally, the motion is uniform (a X =0) so there is only one formula we can use (v=d/t). Ï Vertically, the motion is nonuniform since gravity is influencing the motion (a Y =a g =9.8m/s 2 ). Thus we can use the formulas for uniformly accelerated motion. November 24, U1-10 NOTE! Even though the two motions are independent of one another they do share one common factor: time. The time taken for the horizontal motion is exactly the same as the time taken for the vertical motion. November 24, U1-11 4

5 Problems PROJECTILE MOTION PROBLEMS the motion is 2D (the horizontal and vertical motions are independent) the horizontal motion (x-axis) is uniform a x = 0 L usev x = d x / t x the vertical motion (y-axis) is non-uniform a y = a g = 9.8 m/s 2 [d] L use5 uniform acceleration formulas time is common to both t x = t y November 24, U1-12 Launching a Projectile Horizontally One of the simplest types of projectile motion is when an object is projected horizontally from a known height. In this case, it has an initial velocity in the horizontal direction (v ix ) but the initial velocity in the vertical direction (v iy ) is zero. NOTE! With this type of problem, take motion to the right and down as positive. November 24, U1-13 Launching a Projectile Horizontally 3. A beanbag is thrown from a window located 10.0 m above the ground with an initial horizontal velocity of 3.0 m/s (a) How long will it take the beanbag to reach the ground? That is, what is its time of flight? (a) t = 1.4 s November 24, U1-14 5

6 Launching a Projectile Horizontally 3. A beanbag is thrown from a window located 10.0 m above the ground with an initial horizontal velocity of 3.0 m/s (b) How far will the beanbag travel horizontally? That is, what is its range? (b) d x = 4.3 m November 24, U1-15 Launching a Projectile Horizontally 4. A hockey puck is launched horizontally from the roof of a 32 m tall building at a velocity of 8.6 m/s. (a) What is the hockey puck s time of flight? (a) t = 2.6 s November 24, U1-16 Launching a Projectile Horizontally 4. A hockey puck is launched horizontally from the roof of a 32 m tall building at a velocity of 8.6 m/s. (b) What is the hockey puck s range? (b) d x = 22 m November 24, U1-17 6

7 Launching a Projectile DYK? You can increase the range and time of flight of a projectile by projecting it partially upward instead of horizontally. Since the projectile is moving horizontally at a constant velocity, increasing the time of flight can increase the horizontal displacement. However, if you choose too large an angle, the range of the projectile actually decreases. November 24, U1-18 Launching a Projectile at an Angle 5. Assuming there is no air resistance, at what angle should you launch a projectile from the ground so that it has the greatest time of flight? greatest range? in both cases use 45E November 24, U1-19 Launching a Projectile at an Angle 6. A field hockey ball is launched from the ground at an angle to the horizontal. What are the ball s horizontal and vertical accelerations: (a) at its maximum height? (b) halfway up to its maximum height? (c) halfway down to the ground? (a) a horizontal = 0, a vertical = 9.8 m/s 2 [down] (b) a horizontal = 0, a vertical = 9.8 m/s 2 [down] (c) a horizontal = 0, a vertical = 9.8 m/s 2 [down] November 24, U1-20 7

8 Launching a Projectile at an Angle Projectiles launched at an angle to the horizontal also undergo parabolic motion. The calculations are similar to those you have already done except that now the projectile has an initial velocity in the vertical direction. In this case, we will first need to break the initial velocity into its horizontal and vertical components. NOTE! With this type of problem, take motion to the right and up as positive. November 24, U A soccer player running on a level playing field kicks a soccer ball with a velocity of 9.4 m/s at an angle of 40E above the horizontal. If the ball lands at same level from which it was shot, determine the soccer ball s: (a) initial velocity components. vix = cosθ v i v = v cosθ ix i viy = sinθ v i v = v sinθ iy i November 24, U A soccer player running on a level playing field kicks a soccer ball with a velocity of 9.4 m/s at an angle of 40E above the horizontal. If the ball lands at same level from which it was shot, determine the soccer ball s: (b) time of flight. (b) t = 1.2 s November 24, U1-23 8

9 7. A soccer player running on a level playing field kicks a soccer ball with a velocity of 9.4 m/s at an angle of 40E above the horizontal. If the ball lands at same level from which it was shot, determine the soccer ball s: (c) range. (c) d x = 8.9 m November 24, U A soccer player running on a level playing field kicks a soccer ball with a velocity of 9.4 m/s at an angle of 40E above the horizontal. If the ball lands at same level from which it was shot, determine the soccer ball s: (d) maximum height. (d) d y = 1.9 m November 24, U1-25 There are a set of formulas that have been derived especially for projectile motion. However, these formulas can only be used when the projectile lands at the same level from which it is launched. They are: Range Maximum Height Time of Flight 2 v sin2θ R = g H max = ( vsinθ ) 2g 2 2vsinθ t flight = g November 24, U1-26 9

10 The most complex type of projectile motion problem combines the previous two problem types. In this situation the projectile is launched at an angle from a height above the ground and lands at another height. NOTE! Again, you break the motion into its horizontal and vertical motions and then analyze each separately. However, these situations will not be studied at this level. November 24, U1-27 U Check Your Learning TEXTBOOK P.81 Q.1,3-5 WIKI(KINEMATICS) O... 3U1 - ASG#3 () O... 3U1 - QUIZ#3 (Motion in 2D) November 24, U