Question

A ball rolls off the edge of a table top with an initial speed of 37.0 m/s. In 0.43 s, how far does it travel (a) vertically (b) horizontally?

Answer 1

This question involves the concept of the equations of motion.

(a) vertically the ball travels "0.91 m".

(b) horizontally the ball travels "15.91 m".

(a) Vertical Distance

Applying the second equation of motion in the vertical direction:

$h=v_it+\frac{1}{2}gt^2$

where,

• h = vertical distance = ?
• vi = initial vertical speed of ball = 0 m/s
• t = time = 0.43 s
• g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = (0\ m/s)(0.43\ s)+\frac{1}{2}(9.81\ m/s^2)(0.43\ s)^2$

h = 0.91 m

(b) Horizontal Distance

Taking air friction as negligible the horizontal motion will be uniform motion. Hence, we use the following equation for horizontal motion:

$s=vt$

where,

• s = horizontal distance = ?
• v = horizontal speed of ball = 37 m/s
• t = time = 0.43 s

Therefore,

$s=(37\ m/s)(0.43\ s)$

h = 15.91 m

Learn more about the equations of motion here:

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