Physics College

A marble rolls off a tabletop 1.15 m high and hits the floor at a point 4 m away from the tables edge in the
horizontal direction.
a. How long is the marble in the air?
b. What is the speed of the marble when it leaves the tables edge?
Im/s
C. What is its speed when it hits the floor?

Answers

Answer 1

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/sVT=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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Answers

Answer:

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Answers

Answer:

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