Sure! Let's solve the problem step-by-step.
Given:
- Initial velocity (v₀) = 20 m/s
- Acceleration due to gravity (g) = 9.81 m/s²
Required:
1. Expression for the maximum height.
2. Expression for the velocity on hitting the ground.
### 1. Maximum Height
When a stone is thrown vertically upwards, it reaches its maximum height when its velocity becomes zero. At this point, the initial velocity has been entirely countered by the acceleration due to gravity. We can use the following kinematic equation to find the time to reach the maximum height ([tex]\( t_{max} \)[/tex]):
[tex]\[ v = v₀ - gt \][/tex]
At the maximum height, the final velocity (v) = 0, so:
[tex]\[ 0 = 20 - 9.81t_{max} \][/tex]
Solving for [tex]\( t_{max} \)[/tex]:
[tex]\[ t_{max} = \frac{20}{9.81} \approx 2.038735983690112 \text{ seconds} \][/tex]
Now, we substitute [tex]\( t_{max} \)[/tex] into the following kinematic equation to find the maximum height (h):
[tex]\[ h = v₀t_{max} - \frac{1}{2}gt_{max}^2 \][/tex]
So, plugging in the given values:
[tex]\[ h = 20 \times 2.038735983690112 - \frac{1}{2} \times 9.81 \times (2.038735983690112)^2 \][/tex]
[tex]\[ h \approx 20.38735983690112 \text{ meters} \][/tex]
### 2. Velocity on Hitting the Ground
When the stone falls back to the ground, it will have the same magnitude of velocity as it had initially but in the opposite direction (downward). This is due to the principle of conservation of energy (neglecting air resistance).
Thus, the velocity on hitting the ground ([tex]\( v_{ground} \)[/tex]) is:
[tex]\[ v_{ground} = -v₀ = -20 \text{ m/s} \][/tex]
### Summary:
1. Maximum Height: The maximum height reached by the stone is approximately [tex]\( 20.3874 \text{ meters} \)[/tex].
2. Velocity on Hitting the Ground: The velocity on hitting the ground is [tex]\( -20 \text{ m/s} \)[/tex].
