Question

A mass falls x meters in t seconds. The same mass falls the last half this distance in 1 sec. What is t?

Group of answer choices

2.42 s

6.84 s

3.41 s

0.505 s

Answer 1

Under uniform acceleration, s=ut+1/2 at2, where s is displacement, u is initial velocity, a is acceleration and t is time.

For the given problem, u= 0 m/s (the mass starts at rest), a=9.8 m/s2,s=x, time =t

So, x=0*t+1/2*9.8*t*t

=>x=4.9*t^2...........eqn 1

Also, if in last 1 second, the mass covers half the distnace(=x/2), then in (t-1) seconds, distnace covered = x-x/2=x/2

So, for (t-1) seconds,x/2=1/2*9.8*(t-1)^2

=>x/2=4.9*(t-1)^2

=>x=9.8*(t-1)^2..............eqn 2

From eqn 1 and eqn 2, we get: 4.9*t^2=9.8*(t-1)^2

=>t^2 = 2(t-1)^2

=>t=(2^0.5)*(t-1)

=>t=1.414(t-1)

=>t=1.414t - 1.414

=>0.414 t= 1.414

=>t=3.41 s.

So, third option is correct.