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Sagot :
A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the edge of the table in the horizontal direction,
- t= 0.45 seconds.
- V=2.22m/s
- VT=4.95 m/s
This is further explained below.
What is its speed when it hits the floor...?
Generally, the equation for motion is mathematically given as
S= ut + 0.5at²
Therefore
y = Voy t + 0.5gt^2
1 = 0.5x 98 x 6²
1=4.9t^2
[tex]t=\sqrt{0.2041 }[/tex]
t= 0.45 seconds.
b) Horizontal motions are uniform.
V=Horizontal displacement/time
V=1/0.45
V=2.22m/s
C)
Vx: 2.22 m/s At bottom,
Vy² = Voy² + 2as
Vy² = 2x95x1
Vy² = 19.6
Total velocity
[tex]VT=\sqrt{( 2.22 m/)^2+19.6}[/tex]
VT=4.95 m/s
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