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Unformatted text preview: homework 04 Due: Feb 14 2007, 4:00 am 1 Question 1, chap 1, sect 1. part 1 of 3 5 points A projectile is shot on level ground with a horizontal velocity of 12 m / s and a vertical velocity of 22 m / s . The acceleration of gravity is 9 . 8 m / s 2 . x y v 12 m / s 22m / s h R Find the travel time of the projectile t total . Correct answer: 4 . 4898 s (tolerance 1 %). Explanation: Let : v x = 12 m / s and v y = 22 m / s . t total = 2 t rise = 2 v y g = 2 parenleftbigg 22 m / s 9 . 8 m / s 2 parenrightbigg = 4 . 4898 s . Question 2, chap 1, sect 1. part 2 of 3 5 points Find the range R . Correct answer: 53 . 8775 m (tolerance 1 %). Explanation: R = v x t total = (12 m / s) (4 . 4898 s) = 53 . 8775 m . Question 3, chap 1, sect 1. part 3 of 3 7 points Now consider a new situation, where the magnitude of v is fixed to be the same value as that in part 1 and part 2, and we vary the angle to get the maximum range. Find the height when the range is maxi mum. Correct answer: 16 . 0204 m (tolerance 1 %). Explanation: The magnitude of the velocity is v = radicalBig v 2 x + v 2 y = 25 . 0599 m / s . When the range is maximum, the angle = 45 . Therefore the ycomponent of the velocity is v y = v sin = v 2 . In turn, the height when the range is maxi mum is given by h = v 2 2 a , in our present case h = v 2 2 2 g = (25 . 0599 m / s) 2 4 (9 . 8 m / s 2 ) = 16 . 0204 m . Question 4, chap 4, sect 4. part 1 of 1 7 points A man can throw a ball a maximum hori zontal distance of 153 m on a level field. The acceleration of gravity is 9 . 8 m / s 2 . How far can he throw the same ball verti cally upward from the ground? (Assume that the ball is thrown from the ground and that his muscles give the ball the same speed in each case.) Correct answer: 76 . 5 m (tolerance 1 %). homework 04 Due: Feb 14 2007, 4:00 am 2 Explanation: The range of the ball is given by the expres sion R = v 2 sin2 g The maximum horizontal distance is obtained when the ball is thrown at an angle = 45 and sin2 = 1. Solving for v , v = radicalbig gR When the ball is thrown upward with this speed, the maximum height is obtained from the equation v 2 f = v 2 2 gh Let v f = 0, and solve for h . h = v 2 2 g Question 5, chap 4, sect 4. part 1 of 1 10 points A 0 . 98 kg rock is projected from the edge of the top of a building with an initial velocity of 11 . 3 m / s at an angle 35 above the horizontal. Due to gravity, the rock strikes the ground at a horizontal distance of 20 . 3 m from the base of the building. Assume: The ground is level and that the side of the building is vertical. The accelera tion of gravity is 9 . 8 m / s 2 ....
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 Spring '08
 Turner
 Acceleration, Gravity, Work

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