Question

A ball is dropped from the top a a building and takes a time of 0.5 seconds to pass 3 metre length of a window,which is at some distance from the top of a building. If velocities of the ball at the top and bottom of the building are V(t) and V(b) respectively, find 1: V(t)+V(b) 2:V(t) 3:V(b) please answer fast for 15 points ​

Answer 1

Answer:

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Explanation:

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 VT and VB respectively then (g=9.8m/s^2 As acceleration due to gravity  g=9.8m/s^2 We can write VB=VT+ gt => VB- VT = 9.8x0.5 =5=4.9m/s .....(1) VB^2-VT^2=2*g*3=2*9.8*3.....(2) Dividing (2) by(1) we get VB+VT =(2*9.8*3)/(4.9)=12m/s Hence correct option is (A)