Please answer questions 004,005,006,007 only Thanks This print-out should have 9

Question

Please answer questions 004,005,006,007 only

Thanks

This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. A particle mows in the xy plane with constant acceleration. At time zero, the particle is at x = 1.5 m, y = 4 m, and has velocity r_o = (2.5 m/s) i + (-4.5 m/s) j. The acceleration la given by a = (6.5 m/s^2) i + (6.5 m/s^2) j. What is the x component of velocity after 2.5 s? Answer in units of m/s. What is the y component of velocity after 2.5 s? Answer in units of m/s. What in the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 2.5 s? Answer in units of m. Neglect: Air friction: Your teacher tosses a basketball. The hall gets through the hoop (lucky shot). How long does it take the ball to reach its maximum height? Answer in units of s. How long does it take the ball to reach the hoop? Answer in units of s. What is the horizontal length t of the shot? Answer in units of m. A cat chases a mouse across a 1.3 m high table. The moose steps out of the way, and the cat slides off the table and strikes the floor 1.7 m from the edge of the table. What was the cat's speed when it slid off the table? The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s. A particle travels horizontally between two parallel walls separated by 18.4 m. It moves toward the opposing wall at a constant rate of 5.3 m/s. Also, it has an acceleration in the direction parallel to the walls of 1.9 m/s^2 What will be its speed when it hits the opposing wall? Answer in units of m/s. At what angle with the wall will the particle strike? Answer in units of

Explanation / Answer

4] At maximum height, vy = 0

0 = uy - gt

=> t = uy/g = (17sin65o)/9.8 = 1.572 s.

5] (h - h') = uyt - (1/2)gt2

=> 3.048 - 2.702 = 17sin65ot - (1/2)9.8t2

=> 4.9t2 - 15.407t + 0.346 = 0

this is a quadratic equation in t whose solution is:

t = 0.0226s and t = 3.1216 s

we can discard the first value since we know that the highest point of the trajectory was at t = 1.572s.

therefore the time taken to reach the hoop = 3.1216s.

6] l = uxT = (17cos65o)3.1216 = 22.43 m

7] uy = 0

and h = uyt - (1/2)gt2

=> -1.3 = 0 - (1/2)9.81t2

=> t = 0.515s

L = 1.7 m = uxt

=> ux = L/t = 1.7/0.515 = 3.3 m/s

therefore the initial speed of the cat was 3.3 m/s.

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