# A ball rolls off of a table 1.45m high. The ball lands 1.21 from the base of the table.  a) How much time does the ball take to hit the floor?  b) How fast was the ball going when it rolled off the table?  c) What is the velocity with which the ball hits the floor? (NotZero!!) Give magnitude and direction. Given information: Height of table = H = 1.45 m

Horizontal distance traveled, R = 1.21 m

At initial instant, there is no vertical velocity component (i.e., `u_y = 0.` ). The velocity has only the horizontal component, `u_x` .

Using the equation of motion:

`H = u_yt + 1/2 at^2`

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Given information: Height of table = H = 1.45 m

Horizontal distance traveled, R = 1.21 m

At initial instant, there is no vertical velocity component (i.e., `u_y = 0.` ). The velocity has only the horizontal component, `u_x` .

Using the equation of motion:

`H = u_yt + 1/2 at^2`

here, t is time of travel, and the only acceleration is due to the gravity (= 9.81 m/s^2). We can use a sign convention of negative for downward motion, which means H = -1.45 m and a = -g = -9.81 m/s^2

Substituting these values in the equation, we get:

`-1.45 = (0)xxt + 1/2 (-9.81)t^2`

`or, t = sqrt[(2(-1.45))/(-9.81)] = 0.54 sec`

Using the equation of motion for horizontal distance and noting that there is no horizontal component of acceleration (gravity is only in downward direction), so the horizontal velocity will stay the same throughout the motion.

thus, `R = u_x t`

or, `u_x = R/t = 1.21/0.54 m/s = 2.24 m/s`

The vertical component of the velocity when the ball hits the ground can be calculated as:

`v_y = v_i + at = 0 + (-9.81)xx(0.54) m/s = -5.30 m/s`

The horizontal component of the final velocity is same as the initial velocity.

Therefore, the magnitude of the final velocity can be calculated as:

`v = sqrt(v_x^2 + v_y^2) = sqrt ((2.24)^2 + (-5.30)^2) = 5.75 m/s`

The direction of the velocity (with respect to horizontal) can be calculated as:

`theta = tan^(-1) [V_y/V_x] = tan^(-1)[5.30/2.24] = 67.08` degrees

Hope this helps.

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