Answer:
x = 5[km]
Explanation:
We must convert the time from minutes to hours.
![t=30[min]*\frac{1h}{60min}= 0.5[h]\\](https://tex.z-dn.net/?f=t%3D30%5Bmin%5D%2A%5Cfrac%7B1h%7D%7B60min%7D%3D%200.5%5Bh%5D%5C%5C)
We know that speed is defined as the relationship between space and time.

where:
x = space [m]
t = time = 0.5 [h]
v = velocity [m/s]
Now replacing:
![x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]](https://tex.z-dn.net/?f=x%20%3D%2010%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A0.5%5Bh%5D%5C%5Cx%3D5%5Bkm%5D)
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.
Part A)
Height reached, y = a*[t^2] / 2 = 58.8 m/s^2 * [7.00 s]^2 / 2 = 1440.6 m
Now you need the final speed to use it as initial speed of the next part.
Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s
Part B) Free fall
Maximum height, y max ==> Vf = 0
Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s
ymax = yo + Vo*t - g[t^2] / 2
ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m
Answer: ymax = 10084.2m
Answer:
4m/s
Explanation:
May be different considering how long the pole is and how heavy the firefighter is.
Answer:
acceleration...............