Answer:
16.4 s
Explanation:
For AB
initial velocity, u = 0
final velocity, v = ?
acceleration, a = 9.8 m/s^2
distance, s = 62 m
Let the time taken is t.
Use second equation of motion
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]62=0+0.5\times9.8\times t^{2}[/tex]
t = 3.56 s
Use first equation of motion
v = u + at
v = 0 - 9.8 x 3.56
v = - 34.89 m/s
For BC
Let the time taken is t'
initial velocity, v = 34.89 m/s
final velocity, v' = 2.8 m/s
a = 2.5 m/s^2
use first equation of motion
v' = v + a t'
2.8 = 34.89 - 2.5 t'
t' = 12.84 s
So, the total time taken
T = t + t' = 3.56 + 12.84 = 16.4 s
