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Oxana [17]
3 years ago
7

A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 3 m and it

does so in 0.49 s. The ball accelerates all the way down; let vA be its speed as it passes the window’s top A and vB its speed as it passes the window’s bottom B. O A B 3 m bb b b b b b b b b b b x y How much did the ball speed up as it passed the window; i.e., calculate ∆vdown = vB −vA ? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s. 017 (part 2 of 2) 10.0 points Calculate the speed vA at which the ball passes the window’s top. Answer in units of m/s.
Physics
1 answer:
Usimov [2.4K]3 years ago
3 0

Answer:

Part a)

\Delta v_{down} = 4.8 m/s

Part b)

v_a = 3.71 m/s

Explanation:

Since ball is dropped under uniform gravity

so here we can say that the distance of 3 m moved by the ball under uniform acceleration is given as

d = (\frac{v_f + v_i}{2})t

so we have

3 = (\frac{v_f + v_i}{2})(0.49)

v_f + v_i = 12.24

also we know that

v_f - v_i = at

v_f - v_i = (9.81)(0.49)

v_f - v_i = 4.81

now we will have

v_f = 8.52 m/s

v_i = 3.71 m/s

Part a)

\Delta v_{down} = v_b - v_a

\Delta v_{down} = 8.52 - 3.71

\Delta v_{down} = 4.8 m/s

Part b)

speed at the top of the window is

v_a = 3.71 m/s

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