Respuesta :
To solve this problem, we'll use the equations of motion under gravity:
i. To find the initial velocity of the bullet:
[tex]\[v = u + gt\][/tex]
ii. To find the total distance traveled by the bullet:
[tex]\[s = ut + \frac{1}{2}gt^2\][/tex]
iii. To find the time taken for the bullet to hit the ground:
[tex]\[s = ut + \frac{1}{2}gt^2\][/tex]
Given:
- Initial height (u) = 2m
- Acceleration due to gravity (g) = 10 m/s²
- Time to reach maximum height (t) = 4s
We can solve these equations step by step:
i. Initial velocity (u):
[tex]\[v = u + gt\]At maximum height, the velocity becomes 0.So, at maximum height, \(v = 0\).\[0 = u + g \times 4\]\[u = -40 m/s\][/tex]
ii. Total distance traveled by the bullet:
[tex]\[s = ut + \frac{1}{2}gt^2\]At maximum height, the displacement is equal to the initial height.\[2 = -40 \times 4 + \frac{1}{2} \times 10 \times 4^2\]\[2 = -160 + 80\]\[2 = -80 + 80\]\[2 = 0\][/tex]
This doesn't seem right. We may have made a mistake. Let's reassess.
The correct approach for ii is to calculate the total distance traveled during ascent and descent separately.
For ascent:
[tex]\[v = u + gt\]At maximum height, \(v = 0\).\[0 = u + 10 \times 4\]\[u = -40 m/s\][/tex]
Now, let's calculate the distance traveled during ascent:
[tex]\[s_{\text{ascent}} = ut + \frac{1}{2}gt^2\]\[s_{\text{ascent}} = -40 \times 4 + \frac{1}{2} \times 10 \times 4^2\]\[s_{\text{ascent}} = -160 + 80\]\[s_{\text{ascent}} = -80\][/tex]
For descent:
The bullet will travel the same distance down as it did up, so we don't need to recalculate it. We just need to consider the initial height.
[tex]\[s_{\text{descent}} = 2 \text{ meters}\][/tex]
The total distance traveled will be the sum of the distances during ascent and descent:
[tex]\[s_{\text{total}} = |s_{\text{ascent}}| + s_{\text{descent}}\]\[s_{\text{total}} = 80 + 2\]\[s_{\text{total}} = 82 \text{ meters}\][/tex]
iii. Time taken for the bullet to hit the ground:
The total time is twice the time taken to reach maximum height since the motion is symmetrical.
[tex]\[t_{\text{total}} = 2 \times 4\]\\\[t_{\text{total}} = 8 \text{ seconds}\][/tex]
So, the solutions are:
[tex]i. The initial velocity of the bullet is \( -40 \, \text{m/s}\).\\ii. The total distance traveled by the bullet by the time it hits the ground is \(82 \, \text{meters}\).\\iii. The time taken for the bullet to hit the ground is \(8 \, \text{seconds}\).[/tex]
