Question

solve with steps explanation ✅☑✔​

Answer 1

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

Answer 2

Answer:

Height of tower= displacement of an object= average velocity × time=5×5=25m,

Let initial velocity of the object is u.

Using s=ut+

2

1

at

2

, ⇒−25=u×5−

2

1

g×5

2

⇒u=20ms

−1

,

Now from the tower,Let maximum height reached by object is h.

At maximum height velocity v=0,

Using v

2

=u

2

+2as ⇒0=20

2

−2gh ⇒h=20m,

∴ Total distance =2h+25=65m

Average speed=

Total Time

Total Distance

,

⇒ Average speed=

5

65

=13ms

−1