Question

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the
building is 60 m above the ground. How much time elapses between the instant of release and the instant
of impact with the ground?

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Answer 1

2 sec

Explanation:

S=ut+1/2at^2

60=20t+1/2x10t^2

60=20t+5t^2

5t^2+20t-60=0

t^2+4t-12=0

t^2+6t-2t-12=0

t(t+6)-2(t+6)=0

(t+6)(t-2)=0

Since time cannot be =0

Hence t=2sec