Question

Please solve this question step by step ​

Answer 1

Step-by-step explanation:

The ball is dropped from a point A at a height h above the surface. Thus during the entire motion, let the time taken by the ball to move move from A to B be t¹ from B to C and C to B be t ² , from C to D be t³ , from D to C be t ⁴ , and from B to A be t5

Let T be the time period of the complete motion.

Now for the motion A to B; h=0+ $\frac{1}{2}$g(t¹) ²

⟹ t ¹ = $\frac{√2h}{g}$

Similarly , $t ³$= $t ⁴$= $t 5$= $t ¹$ = $\frac{√2h}{g}$

As the time period of the particle executing SHM inside the tunnel is given by t=2π $\drac{√Re}{g}$

Thus the time period of the motion of the ball inside the tunnel i.e B to C and C to B, $t ²$ =2π $\frac{√R}{g}$

Now the total time period of the motion T= $t ¹$+ $t²$+ $t³+[tex]t⁴$+ $t5$

==> T = 2π $\frac{√R}{g}$ + 4 $\frac{√2h}{g}$

Answer 2

Step-by-step explanation:

is answer hope it helps to you