Question

SOOMS!
An object is dropped from the top of the tower of height 156.8 m and at the same time
another object is thrown vertically upward with the velocity of 78.1 m/s from the foot of
the tower, when and where the objects meet? [ Ans: 2 sec, and 20 m from the top of tower]

Answer 1

Answer:

please mark as brainleast

Explanation:

Solution:

Given height of tower (h) = 156.8 m

Two objects meet at distance x from foot at time t

i). For free fall object:

h = u  . t +   gt2        

156.8 –x =  0 +     X 9.8 X t2                                        

156.8 –x =  4.9 t2              ……………………………… (1)

ii). For vertically upward projected object:

Initial velocity (u) = 78.1 m/s

h = u  . t -   gt2        

x = 78.1  X  t -   X 9.8 X t2                                        

x - 78.1  t = - 4.9 t2  

78.1  t – x = 4.9 t2  …………………………………………….. (2)

Now, eq. (1) and (2) we get

78.1  t – x = 156.8 –x

∴ t =   = 2.01 sec

Also, using value of t in eq. (1), we get,

156.8 –x =  4.9  X ( 2.01 )2      

x =       156.8 – 19.79  

x = 137.01 m                        ( from foot)

⇒  156.8 –x = 19.79  m                     ( from top of a tower)

Thus, two objects meet at 2.01 sec at the distance 137.01m from foot or 19.79m from top of tower.

Answer 2

Answer:

The two objects meet after 2 sec at a distance of 20.6m from the top of the tower

Explanation:

Given that,

An object is dropped from the top of the tower of height 156.8 m and at the same time

another object is thrown vertically upward with the velocity of 78.1 m/s from the foot of

the tower.We are required to find when and where the objects meet.

Let us assume that the objects meet at a height y from the foot of the tower at time t seconds.

Applying equation of motion for freely falling object,we get

$156.8 - h = ut - \frac{1}{2} g {t}^{2} \\ = > 156.8 - h = 0 - \frac{1}{2} \times 10 \times {t}^{2} \\ = > 156.8 - h = 5 {t}^{2} - > (1)$

Similarly using the equation for the vertically projected object,

$h = 78.1 t - 5 {t}^{2} - - > (2)$

Substituting equation (2) in equation(1),we get

$156.8 - (78.1t - 5 {t}^{2} ) = 5 {t}^{2} \\ = > 156.8 - 78.1t = 0 \\ = > t = \frac{156.8}{78.1} = 2.007 \: s≈2s$

Substituting this value of time in equation (2) we get,

$y = 78.1(2) - 5(2) {}^{2} = 136.2m$

From the top of the tower the distance becomes

$156.8 - 136.2 = 20.6m$

Therefore,The two objects meet after 2 sec at a distance of 20.6m from the top of the tower

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