Question

if a free falling body in the last second travels a distance equal to first 3 seconds what is totla time ​

Answer 1

Explanation:

Distance travelled in first 3 seconds=(1/2)gt

2

=(1/2)×9.8×9=44.1m

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−0.5)=44.1;

hencet=

9.8

44.1

+0.5=5s