Respuesta :
Answer:
Speed at which the ball passes the window’s top = 10.89 m/s
Explanation:
Height of window = 3.3 m
Time took to cover window = 0.27 s
Initial velocity, u = 0m/s
We have equation of motion s = ut + 0.5at²
For the top of window (position A)
[tex]s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2[/tex]
For the bottom of window (position B)
[tex]s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2[/tex]
[tex]\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673[/tex]
We also have
[tex]t_B-t_A=0.27[/tex]
Solving
[tex]t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s[/tex]
So after 1.11 seconds ball reaches at top of window,
We have equation of motion v = u + at
[tex]v_A=0+9.81\times 1.11=10.89m/s[/tex]
Speed at which the ball passes the window’s top = 10.89 m/s
