Kaplunovsky - hw 04

# Kaplunovsky - hw 04 - homework 04 – – Due 4:00 am 1...

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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 θ 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 θ 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 θ = 45 ◦ . Therefore the y-component of the velocity is v y = v sin θ = 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 θ g The maximum horizontal distance is obtained when the ball is thrown at an angle θ = 45 ◦ and sin2 θ = 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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Kaplunovsky - hw 04 - homework 04 – – Due 4:00 am 1...

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