Answer:
a) h = 593.50 m
b) h₁₁ = 103 m
c) vf = 107.91 m/s
Explanation:
a)
We will use second equation of motion to find the height:
[tex]h = v_{i}t + (\frac{1}{2})gt^2[/tex]
where,
h = height = ?
vi = initial speed = 0 m/s
t = time taken = 11 s
g = 9.81 /s²
Therefore,
[tex]h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\[/tex]
h = 593.50 m
b)
For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:
[tex]v_{f} = v_{i} + gt[/tex]
where,
vf = final velocity at tenth second = v₁₀ = ?
t = 10 s
vi = 0 m/s
Therefore,
[tex]v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s[/tex]
Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:
[tex]h = v_{i}t + (\frac{1}{2})gt^2[/tex]
where,
h = height covered during last second = h₁₁ = ?
vi = v₁₀ = 98.1 m/s
t = 1 s
Therefore,
[tex]h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\[/tex]
h₁₁ = 103 m
c)
Now, we use first equation of motion for complete motion:
[tex]v_{f} = v_{i} + gt[/tex]
where,
vf = final velocity at tenth second = ?
t = 11 s
vi = 0 m/s
Therefore,
[tex]v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)[/tex]
vf = 107.91 m/s
