Solution 46
Let the drag force be proportional to the square of the speed. The deceleration due to air resistance is given by $a_d = c v^2$, where $c$ is a constant.
Step 1: Just Before the Hit
The ball is moving horizontally with speed $u = 20$ m/s.
- Acceleration due to gravity acts vertically downwards: $\vec{g} = -10\hat{j}$.
- Acceleration due to drag acts opposite to velocity (horizontally backwards): $\vec{a}_d = -c u^2 \hat{i}$.
The magnitude of the net acceleration is given as $a = \sqrt{164}$ m/s².
$$ a_{net}^2 = |\vec{g}|^2 + |\vec{a}_d|^2 $$ $$ 164 = (10)^2 + (c u^2)^2 $$ $$ 164 = 100 + (a_d)^2 $$ $$ (a_d)^2 = 64 \implies a_d = 8 \, \text{m/s}^2 $$We can now find the constant coefficient $c$ using $a_d = c u^2$:
$$ 8 = c (20)^2 $$ $$ c = \frac{8}{400} = 0.02 \, \text{m}^{-1} $$Step 2: Just After the Hit
The ball is hit and starts moving vertically upwards with velocity $v = 10$ m/s.
- Gravity still acts downwards: $g = 10$ m/s².
- Drag acts opposite to velocity (downwards). Its new magnitude is $a’_d = c v^2$.
Since both gravity and drag act vertically downwards, the total acceleration is:
$$ a_{total} = g + a’_d $$ $$ a_{total} = 10 + 2 = 12 \, \text{m/s}^2 $$
Answer: The acceleration of the ball immediately after the hit is $12 \text{ m/s}^2$.