Question

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Answer 1

A ball is dropped from a height of $5\mathrm{m}$ onto a sandy floor and penetrates the sand up to $10~\mathrm{cm}$ before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

Solution: In order to find retardation of ball in the sand, it is necessary to find the velocity that was before hit the sand. t1 – time that the ball flies before entering the sand, H (5m) – height from the sand surface, h (10 cm) – distance which ball covered in the sand before stop.

The equation of motion for the ball relative to the X-axis before entering the sand:

$$H = \frac {g t _ {1} ^ {2}}{2}; (V _ {s t a r t} = 0)$$ $$t _ {1} = \sqrt {\frac {2 H}{g}};$$

The rate equation for the ball before entering the sand:

$$V = g t _ {1} = \sqrt {2 g H} (1)$$

The equation of speed after entering the sand:

$0 = V - at_{2}, \quad t_{2} - time$ after which the ball stops completely

$$t _ {2} = \frac {V}{a} = \frac {\sqrt {2 g H}}{a} (2)$$

The equation of motion for the ball after entering the sand:

$$h = V t _ {2} - \frac {a t _ {2} ^ {2}}{2} (3)$$

(1) and (2) to (3): $2h = 2\sqrt{2gH}\frac{\sqrt{2gH}}{a} - \frac{2gH}{a}$ ;

$$2 h = \frac {2 g H}{a};$$ $$a = \frac {g H}{h}.$$

Substitute the numerical values:

$$a = \frac {9 . 8 \frac {m}{s e c ^ {2}} + 5 m}{0 . 1 m} = 4 9 0 \frac {m}{s e c ^ {2}}.$$

Answer: $490\frac{m}{sec^2}$