A ball is dropped from the top of a building. The ball takes 0.5s to f

1.	A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m 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 PLEASE GIVE DETAILED SOLUTION OF THE D3QUESTION IN THE ATTATCHMENT
Answer» A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m height of a window som e distance from top of the building If the speeds are of the ball at top and bottom of the window are V_T and V_Brespectively then (g=9.8m/s^2 As acceleration due to gravity g=9.8m/s^2 We can write V_B=V_T+ gt => V_B- V_T = 9.8x0.5 =5=4.9m/s .....(1) V_B^2-V_T^2=2g3=29.83.....(2) Dividing (2) by(1) we get V_B+V_T =(29.83)/(4.9)=12m/s Hence correct option is (A) We have given, the speed of the ball at the top of the window = V_T and and the speed of the ball at the bottom of the window = V_B 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, V²-U²= 2g(3) also, V = U+g(0.5) i.e V_B-V_T= 0.5g (V_B+V_T)(V_B-V_T) = 6g (V_B+V_T)(0.5g) = 6g V_B+V_T= 12 m/s, V_T-V_B= -4.9 m/s So option (A) is correct i.e V_B+V_T= 12 m/s

A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m 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 PLEASE GIVE DETAILED SOLUTION OF THE D3QUESTION IN THE ATTATCHMENT

Answer»

A ball is dropped from the top of a building. The ball takes 0.5s to fall past the 3m height of a window som e distance from top of the building If the speeds are of the ball at top and bottom of the window are V_T and V_Brespectively then (g=9.8m/s^2

As acceleration due to gravity g=9.8m/s^2

We can write V_B=V_T+ gt

=> V_B- V_T = 9.8x0.5 =5=4.9m/s .....(1)

V_B^2-V_T^2=2*g*3=2*9.8*3.....(2)

Dividing (2) by(1) we get

V_B+V_T =(2*9.8*3)/(4.9)=12m/s

Hence correct option is (A)

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

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

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, V²-U²= 2g(3)

also, V = U+g(0.5) i.e V_B-V_T= 0.5g

(V_B+V_T)(V_B-V_T) = 6g

(V_B+V_T)(0.5g) = 6g

V_B+V_T= 12 m/s,

V_T-V_B= -4.9 m/s

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

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