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 ?

Arun
25758 Points
3 years ago
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