Question

. A freely falling body covers 44.1m in the last second of its journey. The total distance traveled by the body

Answer 1

Answer:

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