To solve this problem, let's break it down into three parts.
### Part a: How long did it take the truck to hit the ground?
We start by determining the time it took for the truck to fall 100 meters vertically. We use the kinematic equation for free fall:
[tex]\[ h_\text{cliff} = \frac{1}{2} g t^2 \][/tex]
Where:
- [tex]\( h_\text{cliff} \)[/tex] is the height of the cliff (100 m)
- [tex]\( g \)[/tex] is the acceleration due to gravity (9.8 m/s²)
- [tex]\( t \)[/tex] is the time in seconds
Rearranging to solve for [tex]\( t \)[/tex]:
[tex]\[ t^2 = \frac{2 h_\text{cliff}}{g} \][/tex]
[tex]\[ t = \sqrt{\frac{2 h_\text{cliff}}{g}} \][/tex]
Plugging in the values:
[tex]\[ t = \sqrt{\frac{2 \times 100}{9.8}} \][/tex]
[tex]\[ t \approx 4.52 \text{ seconds} \][/tex]
### Part b: What was the velocity of the truck as it went over the cliff?
Next, we find the horizontal velocity of the truck. Given that it traveled 50 meters horizontally before hitting the ground, we use:
[tex]\[ d_\text{base} = v_\text{horizontal} \times t \][/tex]
Where:
- [tex]\( d_\text{base} \)[/tex] is the horizontal distance (50 m)
- [tex]\( v_\text{horizontal} \)[/tex] is the horizontal velocity
- [tex]\( t \)[/tex] is the time from part a (4.52 seconds)
Rearranging to solve for [tex]\( v_\text{horizontal} \)[/tex]:
[tex]\[ v_\text{horizontal} = \frac{d_\text{base}}{t} \][/tex]
Plugging in the values:
[tex]\[ v_\text{horizontal} = \frac{50}{4.52} \][/tex]
[tex]\[ v_\text{horizontal} \approx 11.07 \text{ m/s} \][/tex]
### Part c: What was the velocity of the truck on impact?
Finally, we need to find the impact velocity. The truck will have both horizontal and vertical components of velocity at the moment of impact.
1. Horizontal velocity: This remains constant at 11.07 m/s.
2. Vertical velocity: This can be found using [tex]\( v = g \times t \)[/tex]:
[tex]\[ v_\text{vertical} = g \times t \][/tex]
[tex]\[ v_\text{vertical} = 9.8 \times 4.52 \][/tex]
[tex]\[ v_\text{vertical} \approx 44.27 \text{ m/s} \][/tex]
The resultant impact velocity can be found using the Pythagorean theorem:
[tex]\[ v_\text{impact} = \sqrt{v_\text{horizontal}^2 + v_\text{vertical}^2} \][/tex]
Plugging in the values:
[tex]\[ v_\text{impact} = \sqrt{(11.07)^2 + (44.27)^2} \][/tex]
[tex]\[ v_\text{impact} \approx 45.63 \text{ m/s} \][/tex]
### Summary of Results
- Time to hit the ground (t): 4.52 seconds
- Horizontal velocity as it went over the cliff (v_horizontal): 11.07 m/s
- Vertical velocity on impact (v_vertical): 44.27 m/s
- Velocity on impact (v_impact): 45.63 m/s
