Answer to Question #4450 in Mechanics | Relativity for archana
Suppose the pebble had initial velocities Vx and Vy (V=sqrt(Vx^2+Vy^2)). As pebbles hit the window with only a horizontal component of velocity Vx, so
the vertical component becomes equal to zero while the pebble is in the air.
Let’s find this time t from the equation for vertical displacement:
H = 4.5 = (gt^2)/2 => t = 3/sqrt(g).
The equation for horizontal displacement is:
S = 5 = Vx*t => Vx = 5/t = 5*sqrt(g)/3 = 5*sqrt(9.8)/3 = 5.22 m/s.
So, Vx = 5.22 m/s.
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