Question

 A ball is dropped from the edge of a roof. it takes 0.1 s to cross a window of height 2.0m. find the height of the roof above the top of the window ______________

(a) 1.96m

(b) 196m

(c) 96m

(d) 19.6m

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Answer 1

Answer:

1.96

Explanation:

as the ball will fall so we will multiply the 0.1 and 2.0m as ball us at the edge of the window then we will divide 1.96 by 0.1 so the answer is 1.96

Answer 2

Answer:

Hope it helps you.....(✿◕‿◕✿)

Explanation:

We first consider the movement of ball as it covers the length of the window...

we know that

s2 = ut + (1/2)at2

now,

let the initial velocity be v1

time taken, t = 0.1s

distance, s2 = 2m

acceleration, a = g = 9.81 m/s2

so,

2 = 0.1v1 + (1/2)x9.81x0.12

or solving further

0.1v1 = 2 - 0.049

thus, we have

v1 = 19.5 m/s

now,

let us consider the situation when the ball crossed the distance between the edge of the roof and top of the window.

we know that

v2 - u2 = 2as

here,

s is to be determined

u = initial velocity = 0

v = v1 = 19.5 m/s

so,

s1 = [v2 - u2] / 2a

    = [19.52 - 0] / 2x9.81

thus, we get

s1 = 19.4 m

∴ Your answer is d) 19.6 m

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