an object projected vertically up from top of tower took 5 seconds to reach ground . if average velocity of object was 5 m/s , then what is its average speed ?
answer fast guys
Answers
Height of the tower = h m.
Net displacement from initial
position = - h m
Average velocity = - h/5 m/s = - 5 m/s Given
So h = 25 m.
Let the initial speed = u m/s upwards.
Time duration to reach the top = t1
= u/g sec
Height reached above the tower = h2 = u^2 / 2g m
When the object reaches the top of the tower on its way down, it has the same speed as u but downwards. Time taken to cover h = 25m is 5 - 2u/g sec.
=> 25 m = u (5 - 2u/g) + 1/2 g (5 - 2u/g)^2
=> 25 = 5 u - 2 u^2/g + 25 g /2 - 10 u + 2 u^2/g
=> u = 5 (g/2 - 1)
Total Distance travelled from the top of the tower = 2 * u^2
/2g + 25 m
Average speed = (u^2 / g + 25) / 5 = 5 + u^2
/5g m/s
= 5 + 5 (g/2 - 1)^2
/ g
= 5 [ 1 + g/4 + 1/g - 1] = 5 [g/4 + 1/g]
= 5 (2.5 + 0.1) = 13 m/s, for g = 10 m/s^2
Answer: 5 (g/4 + 1/g) or 13 m/s