Amarble rolls off the edge of a table top with a speed of 2.00 m/s. a.) what is the magnitude of its velocity 0.100 s later? b.) how far from the table does it land? the height of the table is 1.00m.

(a) 2.23 m/s

(b) 0.9 m

Explanation:

h = 1 m, t = 0.1 second

horizontal component of initial velocity, ux = 2 m/s

vertical component of initial velocity, uy = 0  

(a) Let v be the velocity after 0.1 seconds. Its vertical component is vy and horizontal component is vx.

The horizontal component of velocity remains constant as in this direction, acceleration is zero.

vx = ux = 2 m/s

Use first equation of motion in Y axis direction.

vy = uy + g t

vy = 0 + 9.8 x 0.1 = 0.98 m/s

Resultant velocity after 0.1 second

v^2 = vx^2 + vy^2

v^2 = 2^2 + 0.98^2

v = 2.23 m/s

(b) Let it takes time t to land.

Use second equation of motion along Y axis

h = uy t + 1/2 g t^2

1 = 0 + 1/2 x 9.8 x t^2

t = 0.45 second

Let it lan at a distance x.

so, x = ux x t

x = 2 x 0.45 = 0.9 m

b.

hope this : )