Question

A ball is dropped from the top of the building .The ball takes 0.5 seconds to fall past the 3 m height of a window some distance from the top of the building .If the speed of the ball at the top and at the bottom of the window are VT and VB respectively,then VT +VB =???

Answer 1

Answer:

$V_T+V_B=20\text{ m/s}$

Explanation:

A ball is dropped from the top of the building .

A ball is dropped from the top of the building .The ball takes 0.5 seconds to fall past the 3 m height of a window some distance from the top of the building

If the speed of the ball at the top and at the bottom of the window are VT and VB

$\text{Speed of the ball at top of window }=V_T$

$\text{Speed of the ball at bottom of window }=V_B$

Using motion equation

$v^2-u^2=2gs$

$v=u+gt$

where,

$v=V_B$

$u=V_T$

$g=9.8\ m/s^2$

$s=3\ m$

$t=0.5\ sec$

Substitute all these value into formula

$V_B^2-V_T^2=2(9.8)(3)=58.8$

$(V_B+V_T)(V_B-V_T)=58.8$------------(1)

$V_B=V_T+(9.8)(0.3)$

$V_B-V_T=2.94$----------(2)

$V_B+V_T=\dfrac{58.8}{2.94}$

$V_T+V_B=20$

Hence, The value of $V_T+V_B=20$ m/s

Answer 2

Answer:

VT +VB = 12 m/s^2

Explanation:

As it is a uniformly accelerated motion,

We know,

average velocity =  total displacement / total time

                              =  $\frac{3m}{0.5 s }$

                              =  6 m/s^1

But,

average velocity =   $\frac{u+v}{2}$

                    6        =  $\frac{VT+VB}{2}$

                 6*2       =  VT+VB

                12 m/s^1 =VT+VB

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