Answer:
31.75 m/s
Explanation:
h = 41.7 m
Let the initial velocity of the second stone is u
Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.
For first stone:
Use second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7
So, 41.7= 0 + 0.5 x 9.8 x t^2
41.7 = 4.9 t^2
t = 2.92 s ..... (1)
For second stone:
Use second equation of motion
[tex]h=ut+\frac{1}{2}gt^2[/tex]
Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity
[tex]h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2[/tex] .... (2)
By equation the equation (1) and (2), we get
[tex]41.7=1.12 u +4.9 \times 1.12^{2}[/tex]
u = 31.75 m/s
