Answer:
The time is [tex]t = 6.17 \ s [/tex]
Explanation:
From the question we are told that
The angle is [tex]\theta = 15^o[/tex]
The height is [tex]h = 12.5 \ m[/tex]
Generally from SOHCAHTOA the length of the ramp can be evaluated as
[tex]L = \frac{h}{sin (\theta)}[/tex]
=> [tex]L = \frac{12.5}{sin (15)}[/tex]
=> [tex]L = 48.3 \ m [/tex]
Generally fro kinematic equation this length is mathematically represented as
[tex]L = ut + \frac{1}{2} * g* sin (\theta )t^2[/tex]
Here u = 0 m/s since the box started from rest
So
[tex] 48.3 = 0 + \frac{1}{2} * 9.8 * (sin(15))* t^2[/tex]
[tex]t = \sqrt{ \frac{2 * 48.3 }{ 9.8 * sin (15)} }[/tex]
[tex]t = 6.17 \ s [/tex]
