Answer:
(a)
vf_1s = 5.19 m/s
h_1s = 10.095 m
vf_4s = 24.23 m/s
h_4s = 4.49 m (below railing)
(b)
vi = 9.9 m/s
(c)
t = 1.53 s
h = 34.41 m
Explanation:
(c)
First, we will use the 1st equation of motion to find the time to attain the highest point:
[tex]v_{f} = v_{i} + gt\\t = \frac{v_{i} - v_{f}}{g}\\[/tex]
where,
t = time to attain maximum height = ?
vf = final velocity = 0 m/s (ball momentarily stops at highest point
vi = initial velocity = 15 m/s
g = - 9.81 m/s (for upward motion)
[tex]t = \frac{0\ m/s - 15\ m/s}{- 9.81\ m/s^2}\\[/tex]
t = 1.53 s
Now, for the height attained we will use the 2nd equation of motion:
[tex]h = v_{i}t + \frac{1}{2}gt^2\\\\h = (15\ m/s)(1.53\ s) + \frac{1}{2}(9.81\ m/s^2)(1.53)^2\\\\h = 22.93\ m + 11.48\ m[/tex]
h = 34.41 m
(b)
using the 3rd equation of motion for a height of 5 m:
[tex]2gh = v_{f}^2 - v_{i}^2\\2(-9.81\ m/s^2)(5\ m) = (0\ m/s)^2 - v_{i}^2 \\v_{i} = \sqrt{98.1\ m^2/ s^2}\\[/tex]
vi = 9.9 m/s
(c)
At t = 1 s:
[tex]v_{f1s} = v_{i} + gt\\v_{f1s} = 15\ m/s + (-9.81\ m/s^2)(1\ s)\\[/tex]
vf_1s = 5.19 m/s
[tex]h_{1s} = v_{i}t + \frac{1}{2} gt^2\\\\h_{1s} = (15\ m/s)(1\ s) + \frac{1}{2}(-9.81\ m/s^2){(1\ s)^2}[/tex]
h_1s = 10.095 m
At t = 4 s:
Since the ball covers the maximum height of 34.41 m in 1.53 s and then starts moving downward.
Therefore for the remaining 4 s - 1.53 s = 2.47 s, the initial velocity will be 0 m/s at the highest point and the value of g will be positive due to downward motion.
[tex]v_{f4s} = v_{i} + gt\\v_{f4s} = 0\ m/s + (9.81\ m/s^2)(2.47\ s)\\[/tex]
vf_4s = 24.23 m/s
[tex]h_{down} = v_{i}t + \frac{1}{2} gt^2\\\\h_{down} = (0\ m/s)(2.47\ s) + \frac{1}{2}(9.81\ m/s^2){(2.47\ s)^2}\\\\h_{down} = 29.92\ m[/tex]
now, for the position with respect to railing:
[tex]h_{4s} = h - h_{down}\\h_{4s} = 34.41\ m - 29.92\ m\\[/tex]
h_4s = 4.49 m (below railing)
