Answer:
v = 17.15 m/s
Explanation:
given,
angle of ramp = 17.0°
length of ramp(l) = 30 m
height of the ramp =
[tex]h = l sin \theta[/tex]
[tex]h = 30 sin 30^0[/tex]
h = 15 m
using energy of conservation
[tex]\dfrac{1}{2}mv^2 = mgh[/tex]
[tex]\dfrac{1}{2}v^2 = gh[/tex]
[tex]v = \sqrt{2gh}[/tex]
[tex]v = \sqrt{2\times 9.8\times 15}[/tex]
[tex]v = \sqrt{294}[/tex]
v = 17.15 m/s
speed of block reaching at the bottom = v = 17.15 m/s
