Question

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

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/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

$t=\sqrt{0.2041 }$

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

$VT=\sqrt{( 2.22 m/)^2+19.6}$

VT=4.95 m/s

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