Answer to Question #59413 in Mechanics | Relativity for muhammad umar

Question

An object is thrown upward from the edge of a tall building with a velocity of 20 m/s. Where will the object be 5 s after it is thrown? take g=10ms-2

a)25 m above the top of the building

b)22 m below the top of the building

c)25 m below the top of the building

d)22 m above the top of the building

Answer 1

the altitude of the object is described by the equation:

y(t) = 20*t - 1/2 * 10 * t^2
or
y(t) = 20 * t - 5 t^2

y(5) = 20*5 - 5*5^2 = 25 - 50 = -25

answer: c)25 m below the top of the building