You want to launch a firework off the roof of a 50 foot building with a initial velocity of 5.4 ft./s so its initial height is 50 feet. you want the firework to explode at 30 feet so it takes it about 2.67 seconds to explode. Create a graph that represents the path of this firework mark your graph to indicate the point at which the firework will explode.
To create a graph representing the path of the firework, we will use the position-time equation for motion in the vertical direction:
h = -16t^2 + vt + s
Where:
h = height of the firework at time t
t = time in seconds
v = initial velocity of the firework (5.4 ft./s)
s = initial height of the firework (50 ft)
To mark the point at which the firework will explode, we will plot a vertical line at t = 2.67 seconds and h = 30 feet.
Now, let's calculate the height of the firework at various time intervals and plot the graph:
When t = 0 seconds,
h = -16(0)^2 + 5.4(0) + 50
h = 50
When t = 0.5 seconds,
h = -16(0.5)^2 + 5.4(0.5) + 50
h = 49.6
When t = 1 second,
h = -16(1)^2 + 5.4(1) + 50
h = 48.4
When t = 1.5 seconds,
h = -16(1.5)^2 + 5.4(1.5) + 50
h = 46.9
When t = 2 seconds,
h = -16(2)^2 + 5.4(2) + 50
h = 44.8
When t = 2.67 seconds (point of explosion),
h = -16(2.67)^2 + 5.4(2.67) + 50
h ≈ 30
Now, let's plot these points on the graph:
```
|
|
| O
| O
| O
| O
| O
| O x
| O
| O
|_______________
0 0.5 1 1.5 2 2.67
```
In the above graph, the "x" represents the point at which the firework will explode.