Question: a ball is projected horizontally from a $45m$ tall cliff with a horizontal velocity of $20ms^{-1}$. How long does it take for the balll to hit the ground
1. **State the problem:**
A ball is projected horizontally from a cliff that is $45m$ tall with a horizontal velocity of $20ms^{-1}$. We need to find the time it takes for the ball to hit the ground.
2. **Relevant formula:**
Since the ball is projected horizontally, the vertical motion is free fall under gravity. The time to fall is given by the formula for vertical displacement:
$$h = \frac{1}{2}gt^2$$
where:
- $h = 45m$ (height of the cliff)
- $g = 9.8ms^{-2}$ (acceleration due to gravity)
- $t$ is the time in seconds
3. **Rearrange the formula to solve for $t$:**
$$t = \sqrt{\frac{2h}{g}}$$
4. **Substitute the known values:**
$$t = \sqrt{\frac{2 \times 45}{9.8}} = \sqrt{\frac{90}{9.8}}$$
5. **Calculate the value inside the square root:**
$$\frac{90}{9.8} \approx 9.1837$$
6. **Find the square root:**
$$t = \sqrt{9.1837} \approx 3.03 \text{ seconds}$$
**Final answer:**
The ball takes approximately $3.03$ seconds to hit the ground.