A particle moves in the xy plane with constant acceleration. At time zero, the p

Question

A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 8 m, y = 9 m, and has velocity v_0^rightarrow = (3.5 m/s) i^hat + (-4 m/s)j^hat. The acceleration is given by a^rightarrow = (2 m/s^2) i^hat + (2.5 m/s^2) j^hat. What is the x component of velocity after 2 s? Answer in units of m/s. What is the y component of velocity after 2 s? Answer in units of m/s. What is the magnitude of the displacement from the origin (x = 0 m, y = 0 m) after 2 s? Answer in units of m. Neglect: Air friction. Your teacher tosses a basketball. The ball 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 l of the shot? Answer in units of m. A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 1.2 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 7.3 m/s. Also, it has an acceleration in the direction parallel to the walls of 3.2 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 degree.

Explanation / Answer

007) h = 1 m ; d = 1.2 m

We know from eqn of motion that,

S = ut + 1/2 a t^2

In the given case, S = h = 1 m ; u = 0 ; a = g = 9.8 m/s^2 ; t = ?

1 = 0.5 x 9.81 x t^2

t = sqrt (1/0.5 x 9.81) = 0.452 s

we know that, v = d/t

v = 1.2/0.452 = 2.655 m/s

Hence, v = 2.65 m/s

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