Question

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.

Answer 1

Part a)

`\Delta v_(down) = 4.8 m/s`

Part b)

`v_a = 3.71 m/s`

Step-by-step 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 = ((v_f + v_i)/(2))t`

so we have

`3 = ((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`