e previous Problem 4.48 Part A You\'re 6.0 m from one wall of a house. You want
ID: 1773615 • Letter: E
Question
e previous Problem 4.48 Part A You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground. Figure 1) Assume the overhang of the roof is negligible, so that you may assume the edge of the roof is 6.0 m from you and 6.0 m from your friend What minimum speed will allow the ball to clear the roof? m/s Submit My Answers Give Up Part B At what angle should you toss the bal? to the horizontal. Submit My Answers Give Up Figure 1of 1 Co 45° 3.0 m 1.0 m m 6.0 m 6.0 mExplanation / Answer
a = -9.8 m/s^2
v(0)= 0
v(t) = da/dt = -9.8t m/s
s(0) = 6m
s(t) = dv/dt = -9.8 / 2t^2 +6 m.
a)
When s(t) = 1 ,
t=sqrt( (6-1)/4.9 )
= 1.01 seconds to fall.
so t for a total duration (up & down) = 2.02 seconds.
The total horizontal distance must be 6+6+6 = 18m
So,
vx = 18/2.02 m/s or 8.909 m/s
a=-9.8 m/s^2
v(t) = da/dt = -9.8t +v(0) m/s
t=1.01
-9.8(1.01 ) +v(0) = 0
v(0)=9.89949 m/s
=> v = sqrt(9.899^2 + 8.909^2)
= 13.32 m/s.
b)
Angle = arctan(8.909/9.899)
=> angle = 48°
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