Question

A body is freely falling. The displacement in the last second is equal to the displacement in the first three
seconds. What is the time of fall?​

Answer 1

Answer:

Let the time of fall be T s. If it falls from height H, then H=(1/2)gT^2………………..(1). In (T-1) second, it covers the distance,

H’=(1/2)g(T-1)^2. Therefore, distance travelled in the last second ,(H-H’)=(1/2)g[T^2-T^2+2T-1]=(1/2)g(2T-1)……………(2).

Now, distance travelled in first 3s is D=(1/2)g(9)…………….(3). But, H-H’=D is given. Then, from(2) and(3), we have,

(1/2)g(2T-1)=(9/2)g. OR 2T-1=9 OR T=5s.

Answer 2

Answer:

"1. 5s"

Explanation:

Distance travelled in first 3 seconds = (1/2)gt2 = (1/2)×9.8×9 = 44.1 m

travelled in first 3 seconds = (1/2)gt2 = (1/2)×9.8×9 = 44.1 m  

travelled in first 3 seconds = (1/2)gt2 = (1/2)×9.8×9 = 44.1 m  Let t be the time taken to travel the total height;

travelled in first 3 seconds = (1/2)gt2 = (1/2)×9.8×9 = 44.1 m  Let t be the time taken to travel the total height;velocity at (t-1)th second = g×(t-1);

distance travelled at the last second = g(t-1)-(1/2)g = g×(t-1.5) = 44.1;

hence t = (44.1/9.8)+0.5 = 5 s.

"Hope this Helpful."