Question

Pls answer this along with its explanation. Thanks in advance!!!

Answer 1

We have given, the speed of the ball at the top of the window = VT and

and the speed of the ball at the bottom of the window = VB

The acceleration we know is ‘g’, g = 9.8 m/s

The me taken to cross the window, t =0.5 s and the length of the window is S = 3 m.

We have, V^2-U^2 = 2g(3)

also, V = U+g(0.5) i.e VB-VT = 0.5g

(VB+VT)(VB-VT) = 6g

(VB+VT)(0.5g) = 6g

VB+VT = 12 m/s,

VT-VB = -4.9 m/s

So option (A) is correct i.e VB+VT = 12 m/s

Answer 2

Answer

Vt + Vb = 12ms^-1

Solution

Given

speed of ball at the top of window = Vt

speed of ball at the bottom of window = Vb

Acceleration = 9.8m/sec^2

T = 0.5 seconds

S = 3m

=> v^2 - u^2 = 2gs

Also, v = u + gt

=> Vb - Vt = 0.5g ..... (1)

put equation (1) in above equation

=> (Vb + Vt ) ( Vb - Vt) = 2g(3)

=> ( Vb + Vt) ( 0.5g) = 6g

=> Vb + Vt = 12 m/s = ANSWER