Answer:
1. t = 5.89 s
2. h = 170 m
3. Vf = 57.8 m/s
Explanation:
1.
First, we analyze the horizontal motion of the golf ball. Assuming the air friction to be negligible, the horizontal motion will be uniform. So, e can use the following equation:
[tex]s = vt[/tex]
where,
s = horizontal distance covered by the golf ball = 471 m
v = horizontal speed of golf ball = 80 m/s
t = time taken by the golf ball in air = ?
Therefore,
[tex]471\ m = (80\ m/s)t\\\\t = \frac{471\ m}{80\ m/s}\\\\[/tex]
t = 5.89 s
2.
Now, we analyze the vertical motion. Using 2nd equation of motion:
[tex]h = v_{i}t + \frac{1}{2}gt^2[/tex]
where,
h = height of cliff = ?
vi = vertical component of initial speed of ball = 0 m/s(ball was shot horizontally)
g = acceleration due to gravity = 9.81 m/s²
t = time of flight = 5.89 s
Therefore,
[tex]h = (0\ m/s)(5.89\ s) + \frac{1}{2}(9.81\ m/s^2)(5.89\ s)^2[/tex]
h = 170 m
3.
Now, we can use 1st equation of motion:
[tex]v_{f} = v_{i} + gt\\v_{f} = 0 m/s + (9.81\ m/s^2)(5.89\ s)\\[/tex]
Vf = 57.8 m/s
