A small marble rolls off the edge of a horizontal table with a horizontal velocity of 3.50 m s^-1. The table top is 0.800 m above the floor. Air resistance is negligible. Take g = 9.81 m s^-2. (a) Show that the time taken for the marble to fall from the table top to the floor is…

AS Level Physics 9702: Projectile Motion of a Marble Leaving a Table

A small marble rolls off the edge of a horizontal table with a horizontal velocity of $3.50\text{ m s}^{-1}$ . The table top is $0.800\text{ m}$ above the floor. Air resistance is negligible. Take $g = 9.81\text{ m s}^{-2}$ .

(a) Show that the time taken for the marble to fall from the table top to the floor is $0.404\text{ s}$ , to 3 significant figures. [3]

(b) Calculate the horizontal distance travelled by the marble, measured from the edge of the table to the point where it lands on the floor. [2]

(d) Calculate the magnitude and direction of the marble's velocity just before it lands on the floor. [3]

Setting up the problem

The key idea in projectile motion is that the horizontal and vertical motions are independent: the horizontal velocity stays constant (no horizontal force, air resistance negligible), while the vertical motion is uniformly accelerated by gravity, starting from a vertical velocity of zero (the marble leaves the table moving purely horizontally).

Part (a): Time of fall

Vertically, $u_y = 0$ , $a = g = 9.81\text{ m s}^{-2}$ , and the marble falls a height $h = 0.800\text{ m}$ . Using $h = u_y t + \tfrac{1}{2}gt^2$ with $u_y = 0$ : $h = \tfrac{1}{2}gt^2 \implies t = \sqrt{\frac{2h}{g}}$ $t = \sqrt{\frac{2(0.800)}{9.81}} = \sqrt{\frac{1.600}{9.81}} = \sqrt{0.16310}$ $t = 0.404\text{ s } (3\text{ s.f.})$ as required.

Part (b): Horizontal distance (range)

Horizontally, the velocity is constant at $v_x = 3.50\text{ m s}^{-1}$ , so: $x = v_x t = 3.50 \times 0.4039$ $x = 1.41\text{ m } (3\text{ s.f.})$

Part (c): Vertical component of velocity on landing

Using $v_y = u_y + gt$ with $u_y = 0$ : $v_y = gt = 9.81 \times 0.4039$ $v_y = 3.96\text{ m s}^{-1} (3\text{ s.f.})$

As a check, using $v_y^2 = u_y^2 + 2gh = 2(9.81)(0.800) = 15.70$ , so $v_y = \sqrt{15.70} = 3.96\text{ m s}^{-1}$ , consistent.

Part (d): Magnitude and direction of the landing velocity

The horizontal component $v_x = 3.50\text{ m s}^{-1}$ and the vertical component $v_y = 3.96\text{ m s}^{-1}$ act at right angles to each other, so they combine using Pythagoras’ theorem: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{(3.50)^2 + (3.96)^2} = \sqrt{12.25 + 15.70}$ $v = \sqrt{27.95} = 5.29\text{ m s}^{-1} (3\text{ s.f.})$

The direction, measured as an angle $\theta$ below the horizontal, is found from: $\tan\theta = \frac{v_y}{v_x} = \frac{3.96}{3.50} = 1.132$ $\theta = 48.5° (3\text{ s.f.})$

Final answers

(a) Time of fall $= \boxed{0.404}\text{ s}$ (shown)
(b) Horizontal distance $= \boxed{1.41}\text{ m}$
(c) Vertical component of velocity $= \boxed{3.96}\text{ m s}^{-1}$
(d) Landing velocity $= \boxed{5.29}\text{ m s}^{-1}$ at $\boxed{48.5°}$ below the horizontal