Question

A book is thrown downward from the library window with a speed of 2.0\,\dfrac{\text m}{\text s}2.0 s m ​ 2, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction and lands on the ground 5.0\, \text m5.0m5, point, 0, start text, m, end text below. We can ignore air resistance. What is the final velocity of the book in \dfrac{\text m}{\text s} s m ​ start fraction, start text, m, end text, divided by, start text, s, end text, end fraction?

Answer 1

Answer: final Velocity v = 10.2m/s

Explanation:

Final speed v(t) is given as

v(t) = u + at .......1

Where; u = the initial speed

a = acceleration

t = time taken

The total distance travelled d is given as

d = ut + 1/2(at^2)

Given

d = 5.0m

u = 2.0m

a = g = 10m/s2 (acceleration due to gravity)

Substituting into the equation above we have

5 = 2t + 5t^2

5t^2 +2t -5 = 0

Applying the quadratic formula. We have;

t = 0.82s & t = -1.22s

t cannot be negative

t = 0.82s

From equation 1 above

v = 2.0m/s + 10(0.82)m/s

v = 10.2m/s

Answer 2

Answer:

-10

Explanation:

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