Presentation on theme: "Projectile Motion Neglecting air resistance, what happens when you throw a ball up from the back of a moving truck? Front? Behind? In? GBS Physics Demo."— Presentation transcript:
Projectile Motion Neglecting air resistance, what happens when you throw a ball up from the back of a moving truck? Front? Behind? In? GBS Physics Demo
Curved Motion Projectile Motion motion of an object that is projected into the earth’s gravitational pull. A projectile will follow the curved path of a parabola, so sometimes it is called parabolic motion. Curved motion is produced by a constant horizontal velocity coupled with an vertical velocity that is accelerating (due to gravity). The affects of air resistance are neglected in order to account for constant horizontal velocity.
An object is thrown from a height of 44.1 m with a horizontal speed of 35.0 m/s. (It will hit in 3 s.) Horizontally: 35 m v e r ti c a ll y 4.9 m 14.7 m 24.5 m
The Monkey and the “Hunter” Problem GBS Physics Demo
Tips for solving Projectile Motion Problems Remember that the curved motion is because the object is moving at a constant rate across while it accelerates due to gravity! The time the object is in the air is independent of the horizontal velocity! How far it falls (or rises) depends upon the vertical components: velocity and the height!
A man throws a ball off of a tower that is 35.0 m tall with a horizontal velocity of 45.0 m/s. How far from the base of the tower will the ball hit? Vertically: v 0 = 0 a = g =-9.80 m/s 2 ∆h = -35.0 m ∆t = ? Horizontally: v h = 45.0 m/s ∆ d = ? ∆d = v h ∆t ∆t = 2∆d a = 2(-35.0 m) -9.80 m/s 2 = 2.67 s = (45.0 m/s)(2.67 s) = 121 m
Sparky the human cannonball sets his cannon so that it will launch him with a velocity of 25.0 m/s at an angle of 35.0˚ with the horizontal. The ceiling of the arena is 10.0 m above his launch point. How far away should he place his landing net? Will Sparky need that net (will he clear the ceiling?) 25.0 m/s 35.0˚ vhvh v Horizontally: v h = (cos 35.0˚)(25.0 m/s) = 20.5 m/s Vertically: v v = (sin 35.0˚)(25.0 m/s) = 14.3 m/s
Horizontally: v h = 20.5 m/s ∆d = ? ∆d = v∆t Vertically: v o = 14.3 m/s v = -14.3 m/s a = g = -9.80 m/s 2 ∆t = ? ∆t = v f - v i a = -14.3 m/s - 14.3 m/s -9.80 m/s 2 = 2.92 s = (20.5 m/s)(2.92 s) = 59.8 m
But does he clear the ceiling? Vertically Upward: ∆t =.5(2.92 s) = 1.46 s v 0 = 14.3 m/s v 0 = 0 ∆d = ? ∆d = v 0 ∆t +.5a∆t 2 = (14.3 m/s)(1.46 s) +.5(-9.80m/s 2 )(1.46 s) 2 = 10.4 m To land, he would have to rise 10.4 m, but the ceiling is only 10.0 m! He will not make it! a = g = -9.80 m/s 2
1) A plane flying horizontally at 352 m/s wants to drop a package so that it lands exactly 1250 m horizontally away from the drop point. At what height should the pilot fly in order to accomplish this? 2) A cannonball is launched with a velocity of 50.0 m/s at an angle of 40.0˚ with the horizontal. What is the range (the farthest horizontal distance) of the cannonball and to what height will it rise?
3) A man throws a baseball horizontally off of the top of a 45.5 m tower and it lands 89.5 m away from the base of the tower. With what velocity did the man throw the ball? 4) A cannonball is fired in such a way that its range (the total horizontal distance traveled) is 325 m and the maximum height it reaches is 65.0 m. What must have been the magnitude and direction of the cannonball’s velocity at its launch?
A projectile if fired at a velocity of 125 m/s @ 35.0˚ from the edge of a 125 m cliff. A) How long until it hits at ground level? B) How far from the base of the cliff will the projectile hit ground? C) What will be the impact velocity of the projectile? A student throws a ball with a velocity of 35.0 m/s @ 65.0˚ at a wall that is 50.0 m high and 59.2 m away from him. Does his throw clear the wall? If not, how high up does hit, and does it hit the wall on the way up or on the way down?