A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 36 m/s when it reaches ground level. What was its speed when its height was half that of its starting point?

its speed when its height was half that of its starting point is 25.46 m/s

Explanation:

Given;

final speed of the roller coaster, v = 36 m/s

Applying general equation of motion;

V² = U² + 2gh

where;

V is the final speed of the roller coaster

U is the initial speed of the roller coaster = 0

h is the height attained at a given velocity

36² = 0 + (2 x 9.8)h

1296 = 19.6 h

h = 1296/19.6

h = 66.1224 m

when its height was half that of its starting point, h₂ = ¹/₂ h

h₂ = ¹/₂(66.1224 m) = 33.061 m

At h = 33.061 m, V = ?

V² = U² + 2gh

V² = 0 + 2 x 9.8 x 33.061

V² = 648

V = √648

V = 25.46 m/s

Therefore, its speed when its height was half that of its starting point is 25.46 m/s

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)