free-fall and sideways motion Projectiles free-fall and sideways motion Chapter 3

Free fall Acceleration is 9.8 m/s2 downward Zero acceleration in the horizontal direction

Board Work What is its horizontal component of velocity? A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. The clam lands on a rocky beach 2.10 s later. Just before the clam lands: What is its horizontal component of velocity? What is its vertical component of velocity? How would your answers change if the crow’s speed were increased? a. Horizontal is always the crow’s speed. b. vertical is (2.10 s)(9.8 m/s2) = 20.58 m/s c. horizontal would increase, vertical would not.

Board Work In Denver, children bring their old Jack-o-lanterns to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower is 9.0 m high and that the bull’s eye is at a horizontal distance of 3.5 m from the tower. If the pumpkin is thrown horizontally, what launch speed is needed to hit the bull’s eye? Time y-y0 = -1/2 gt2 so t2 = 2y0/g = 1.837 s2 and t = 1.355 s, so vx = 3.5 m/1.355 s = 2.582 m/s

Group Board Work A projectile is launched at an initial angle of a0 with a speed of v0. What must a0 be so that the speed at the top of the arc is v0/4? vT = v0x = v0 cos(a0) = v0/4 so cos(a0) = ¼ thus a0 = arccos(1/4) = 75.52°.

Board Work A second baseman tosses the ball to the first baseman, who catches it at the same level at which it was thrown. The throw is made with an initial speed of 17.0 m/s at an angle of 35° above horizontal. What is the ball’s horizontal component of velocity just before it is caught? How long is the ball in the air? How far does it travel horizontally? a. vx = v0x = v0 cos(a0) = (17.0 m/s)(cos 35°) = 13.93 m/s b. 0 = y = v0yt – ½ gt2 s0 t = 0 or 2v0y/g = 2 (17.0 m/s) sin(35°)/(9.8m/s2) = (21 m/s)/(9.8 m/s2) = 1.99 s OR vy = -v0y = v0y-gt so t = 2v0y/g as before. c. x-x0 = vx t = (13.93 m/s)(1.99 s) = 27.70 m

Fun Facts If launch and landing are at the same height, the maximum range is achieved at a launch angle a0 of 45°. Same range is achieved at 45°+q and 45°–q.

Conceptual Question If the landing height is below the launch height, is the launch angle giving the greatest range more or less than 45°? What if the landing height is above the launch height?

Fun Fact For a given v0 and Dy = y–y0, the landing speed v is the same for all a0. Easy proof: use vf2 – v02 = 2 a (y – y0).