Question

Two balls are rolled off a tabletop that is 0.85 m above the floor. Ball A has a

Answer 1

Hi there!

A)
Since the ball's velocity is initially only in the HORIZONTAL direction, there is NO vertical component of its velocity. Therefore, we can treat this like a free-fall scenario. (Dropped from rest).

We can rearrange the following kinematic equation to make solving for the time taken for the balls to hit the ground easier.

$d_y = v_yt + \frac{1}{2}at^2$

dy = vertical displacement (height of table, 0.85 m)

vy = initial vertical velocity (0 m/s, only horizontal)
a = acceleration (due to gravity in this situation, 9.8 m/s²)

t = time (? sec)
Simplify and rearrange the variables for 't'.

$d_y = 0(t) + \frac{1}{2}at^2\\\\d_y = \frac{1}{2}at^2\\\\t^2 = \frac{2d_y}{a}\\\\t = \sqrt{\frac{2d_y}{a}}}$

Plug in the given values.

$t = \sqrt{\frac{2(0.85)}{9.8}} = \boxed{0.4165 s}$

B)

Now, remember that the ball's acceleration in the vertical direction due to gravity does NOT impact its horizontal velocity. Its vertical and horizontal velocities are COMPLETELY independent. Thus, we can simply use the time solved for above and each ball's respective velocities in the following kinematic equation:
$d_x = v_xt$

dₓ= horizontal displacement (? m)

vₓ = horizontal component of velocity (3.5 and 5.3 m/s for A and B respectively)

t = time (0.4165 s)

Solve for the distance traveled by each ball:
Ball A:
$d_x = 3.5 * 0.4165 = \boxed{1.458 m}$

Ball B:
$d_x = 5.3* 0.4165 = \boxed{2.207 m}$