Question

An aluminum ball X and an iron ball Y of the same volume are thrown horizontally with the same velocity from the top of a building. Neglecting air resistance, X reaches the ground (a) before Y and at the same distance from the building (b) at the same time as Y and at a nearer distance to the building (c) at the same time as Y and at the same distance from the building (d) at the same time as Y and further from the building (e) after Y and at the same distance from the building

Answer 1

Explanations & Calculations

• Since the density of Iron is greater than that of Aluminum, mass of the ball Y is greater than X's.

$\qquad\qquad\small m=V\times \rho$

• When they are thrown, they move in projectiles under the gravitational acceleration which is constant throughout the flight of the 2 balls.
• If you closely inspect the 4 equations of motion (these are applicable only for a constant acceleration), $\qquad\qquad \begin{aligned} \small v&=\small u+at\\ \small v^2&=\small u^2+2as\\ \small s&=\small ut+\frac{1}{2}at^2\\ \small s&=\small \frac{(v+u)t}{2} \end{aligned}$ you will see that there is no any influence of the mass of an object over the nature of the flight. Only what happens is $\small a \to g$.

• Therefore, when the influence of air is neglected (in reality we experience the air resistance) both the balls experience the same flight giving (C) as the answer.

• If the height of the building is h & it takes time t for the ball to reach the ground & apply $\small s=ut+\frac{1}{2}at^2$ downward. Then, $\small h=0t_1+\frac{1}{2}gt_1^2 \to t_1=\sqrt{\frac{2h}{g}}$
• Therefore, $\small \bold{t_X=t_Y}$

• If a ball travels some s₁ distance horizontally during that time, apply $\small s= ut+\frac{1}{2}at^2$ for the horizontal motion. Then, $\small s_1 =u_0t_1+\frac{1}{2}\times 0\times t_1\to s_1=u_0t_1 \to s_1=u_0\sqrt{\frac{2h}{g}}$
• Therefore, $\small \bold{s_X=s_Y}$