We have that minimum speed will allow the ball to clear the roof
V=13m/s
From the question we are told that
ou're 6.0 m from one wall of a house
friend who is 6.0 m from the opposite wall.
The throw and catch each occur 1.0 m m above the ground
assume the edge of the roof is 6.0 m from you and 6.0 m from your friend.
Generally the Newtons equation for the distance is mathematically given as
[tex]s = v*t + a/2 * t^2\\\\5 = 0*t + 4.9 * t^2 \\\\ t=1.04[/tex]
Therefore
2*t = 2.0s
Where
[tex]18 = V_x*2.0\\\\V_x=8.9[/tex]
And
[tex]5 = Vy*1.04 - 4.9*1.04^2 \\\\V_y=9.8[/tex]
Therefore
[tex]V=\sqrt{Vy^2+Vx^2}\\\\V=\sqrt{(9.8)^2+(8.9)^2}[/tex]
V=13m/s
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