PROJECTILE MOTION PRACTICE QUESTIONS (WITH ANSWERS) * challenge questions Q1. A golfer practising on a range with an elevated tee 4.9 m above the fairway is able to strike a ball so that it leaves the club with a horizontal velocity of 20 m s-1. (Assume the acceleration due to gravity is 9.80 m s-2, and the effects of air resistance may be ignored unless otherwise stated.) Tee 20 m s-1 4.9 m Fairway a How long after the ball leaves the club will it land on the fairway? b What horizontal distance will the ball travel before striking the fairway? c What is the acceleration of the ball 0.5 s after being hit? d Calculate the speed of the ball 0.80 s after it leaves the club. e With what speed will the ball strike the ground? A1. a x = ut + 0.5at2 then 4.9 m = 0 + 0.5(9.8 m 5-2) +2 and t = 1.0 s b X = (average speed)(time) = (20 m s-1) (1.0 s) = 20 m The acceleration of the ball is constant at any time during its flight, and is equal to the acceleration due to gravity = 9.8 m s-2 down d After 0.80 s, the ball has two components of velocity: Vx = 20 m s-1 and Vy = 0 + (9.8 m s-2)(0.80 s) = 7.84 m s-1 The speed of the ball at 0.80 s is given by: [(20 m s-1)2 + (7.84 m s-1)2] = 21.5 m s-1 e The ball will strike the ground 1.0 s after it is struck. Then Vx = 20 m s-1 and Vy = 0 + (9.8 ms-2) (1.0 s) = 9.8 m s-1 The speed of the ball at 1.0 s is given by: [(20 m s-1)2 + (9.8 m s-1)2] 12 = 22.3 m s-1 Q2. A bowling ball of mass 7.5 kg travelling at 10 m s-1 rolls off a horizontal table 1.0 m high. (Assume the acceleration due to gravity is 9.80 m s-2, and the effects of air resistance may be ignored unless otherwise stated.) a Calculate the ball's horizontal velocity just as it strikes the floor. b What is the vertical velocity of the ball as it strikes the floor? c Calculate the velocity of the ball as it reaches the floor. d What time interval has elapsed between the ball leaving the table and striking the floor? e Calculate the horizontal distance travelled by the ball as it falls. A2. a The horizontal velocity of the ball remains constant and vx = 10 m s-1. b v2 = u2 + 2ax and vy2 = 02 + 2(9.8 m s-2) (1.0 m) and Vy = 4.4 m s-1 down c v = [(10 m s-1)2 + (4.43 m s-1)2] 2 = 10.9 m s-1 at 24° to the horizontal, where the angle is determined from tan 0 = 4.43 m s-1/10 m s-1 = 0.443 and 0 = 24° d x = ut +