Answer:
7.66m/s
Explanation:
Horizontal component of initial speed is
[tex]U_x = Ucos40^0[/tex]
= 10.2 × 0.7660
= 7.66 m/s
horizontal component of the final speed at pinnacle of its trajectory is
[tex]V_x = U_x[/tex]
= 7.66 m/s
the vertical component of the speed at pinnacle of its trajectory is
[tex]V_y = 0[/tex]
the speed of the ball at pinnacle of the trajectory
[tex]V = \sqrt{V_x^2 + V_y^2}[/tex]
= [tex]\sqrt{7.66^2 + 0^2}[/tex]
= 7.66m/s
