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?
Answers
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
Answer:
Initial Velocity (u) = 0 m/s
Acceleration = g = 10 m/s²
Displacement in 3 seconds = Displacement in the last second
To Find:
Total time of free fall by the object.
Solution:
Using the Second Equation of motion we can calculate the displacement in the first 3 seconds. Hence we get,
Now,
Let us assume the last second to be 't'.
We know that, the formula to calculate displacement traversed in a nth second is given as:
Substituting n=t and all the given information we get:
According to the question, (i) is equal to (ii). Hence we get: