University:
Jackson College
Course:
PHY 231 | College Physics I
Academic year:

2022
Views:

263

Pages:

6
Author:

Kayden Benitez

Kinematics #3: Projectiles 1. A golf ball was struck from the first tee at Lunar Golf and Country Club. It was given a velocity of 48 m/s at an angle of 40° to the horizontal. On the moon, g = 1.6 m/s². (a) What are the vertical and horizontal components of the ball's initial velocity? Vx = 48 cos 40° = 36.77 m/s Vy = 48 sin 40° = 30.85 m/s (b) For what interval of time is the ball in flight? t = ? a = -1.6 m/s² Vy = 30.85 m/s Vf = 0 Vf = Vi + at 0 = 30.85 - 1.6t t = 19.28 s (c) How far will the ball travel horizontally? d = vt = (36.77 m/s)(19.28 s) d = 709 m 2. A rock is thrown horizontally from the top of a cliff 98 m high, with a horizontal speed of 27 m/s. (a) For what interval of time is the rock in the air? t = ? d = -98 m a = -9.81 m/s² Viy = 0 d = Vit + 1/2at² -98 = 0 + (-9.81/2)t² t = √(2d/a) = 4.4699 s (b) How far from the base of the cliff does the rock land? dx = Vxt = (27 m/s)(4.4699 s) = 120.67 m (c) With what velocity does the rock hit? Vx = 27 m/s Vyf = ? a = -9.81 m/s² t = 4.4699 s Viy = 0 Vf² = Vi² + 2ad Vf² = 0 + 2(-9.81)(-98) Vf = √(1921.56) = 43.84 m/s tan θ = Vyf/Vx = 43.84/27 = 1.6237 θ = 58° 3. An earthbound golfer strikes a golf ball giving it a velocity of 48 m/s at an angle of 50° to the horizontal. (a) What are the vertical and horizontal components of the ball's initial velocity? Vx = 48 cos 50° = 30.854 m/s Vy = 48 sin 50° = 36.77 m/s (b) How long is the ball in the air? t = ? Vf = 0 Vi = 36.77 m/s a = -9.81 m/s² Vf = Vi + at 0 = 36.77 - 9.81t t = 3.7496 s (c) What is the horizontal distance covered by the ball while in flight? dx = Vxt = (30.854 m/s)(3.7496 s) = 115.69 m (d) What velocity does the ball have at the top of its trajectory? Vx = 31 m/s (only horizontal, which is constant) 4. A rescue pilot wishes to drop a package of emergency supplies so that it lands as close as possible to a target. If the plane travels with a velocity of 81 m/s and is flying 125 m above the target, how far away (horizontally) from the target must the rescue pilot drop the package? t = ? d = 125 m a = -9.81 m/s² Viy = 0 d = Vit + 1/2at² 125 = 0 + (-9.81/2)t² t = √(2d/a) = 5.048 s dx = Vxt = (81 m/s)(5.048 s) = 408.9 m Kinematics #4 1. A ball is thrown with a velocity of 24 m/s at an angle of 30° to the horizontal. (a) What are the vertical and horizontal components of the initial velocity? Vy = 24 sin 30° = 12 m/s Vx = 24 cos 30° = 20.78 m/s (b) How long is the ball in the air? Vf = -Vi Vi = 12 m/s Vf = -12 m/s a = -9.81 m/s² t = (Vf - Vi)/a = 2.4465 s (c) How far away will the ball land? dx = Vxt = (20.7846)(2.4465) d = 50.84 m (d) To what maximum height will the ball rise? d = Vit + 1/2at² d = 12(2.4465) + 1/2(-9.81)(2.4465)² d = 7.3 m (e) With what velocity will the ball land? Vx = 20.78 m/s Vyf = ? a = -9.81 m/s² t = 2.4465 s Viy = 0 Vf² = Vi² + 2ad Vf² = 0 + 2(-9.81)(-7.3) Vf = √143.13 = 11.96 m/s tan θ = Vyf/Vx = 11.96/20.78 = 0.5749 θ = 30° v = 24 m/s [30° below horizontal] 2. A youngster hits a baseball giving it a velocity of 22 m/s at an angle of 62° with the horizontal. How far will the ball travel before a fielder (assuming the fielder catches the ball at the same height that it is hit) catches it? Components: Vy = 22 sin 62° = 19.4242 m/s Vx = 22 cos 62° = 10.3284 m/s time from vertical: t = 3.9602 s dx = Vxt = (10.3284)(3.9602) d = 40.903 m