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Kinematics- Physics


A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of?


18-Nov-2015 2:37 PM
~6

Answers (1)


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Let the total height be H1.

Distance travelled: S = ut +  \frac{1}{2}at2

Distance travelled 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 + \frac{1}{2}aT\frac{1}{2}aT2

(as u = 0)

Distance travelled in (T-1) seconds:

H2 = u(T-1) +  \frac{1}{2}a(T-1)2 = \frac{1}{2}a(T-1)2

(as u = 0)

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

 \frac{1}{2}a(T- (T-1))

\frac{1}{2}a(2T -1)

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

=> 5 - 10T = -45

=> 10T = 50

T = 5 seconds

 

 

 



18-11-2015 15:27
~32

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