Question

A body falls freely from rest. it covers as much distance in the last second if it's motion as covered in the first three seconds. the body has fallen for a time of?

Answer 1

Let the total height be H1.

Distance traveled: S = ut +  1/2 at²

Distance traveled in the first three seconds: (put u=0, a = -10, t = 3 above):

S = - 45m

Let the total time be T. Therefore, distance travelled in complete T seconds:

H1 = uT + 1/2 aT²= 1/2 aT²

(as u = 0)

Distance travelled in (T-1) seconds:

H2 = u(T-1) +  1/2 a(T-1)²= 1/2 a(T-1)²

(as u = 0)

Distance travelled in last second of its motion: H1 - H2

=  1/2 a(T2 - (T-1)2 )

=1/2 a(2T -1)

Putting a = -10 and equating it to (-45) from above:

=> 5 - 10T = -45

=> 10T = 50

T = 5 seconds

Answer 2

Answer: 5 seconds

Explanation:

• Let the total height be H.

• We know  that Distance traveled (S) = $ut+\frac{1}{2}at^{2}$

• Height attained in the first three seconds: (put u=0, a = -10, t = 3 in the above equation) we get S = (- 45) m

• Let the total time taken be T.
• So, total distance travelled in T seconds:

         $H=uT+\frac{1}{2}aT^{2}$

• Distance travelled in (T-1) seconds:

         $H'=u(T-1)+\frac{1}{2}a(T-1)^{2}$

• Distance travelled in last second of its motion:
• (H- H')=  $\frac{1}{2}a[T^{2}-(T-1)^{2} ]$

                   =$\frac{1}{2}a[2T-1 ]$

• Putting a = -10 $m/s^{2}$ and equating it to (-45) m from above:

       ⇒ $\frac{1}{2}(-10)[2T-1 ]=-45$

       ⇒ 10T- 5=45

       ⇒ T=$\frac{50}{10}$

       ∴ T= 5 seconds.

• Hence, the total time taken while falling is 5 seconds.

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