A rock is thrown off a 25 m cliff at 60m/s. It was thrown 32° above the ground How long will the rock be in the air before it hits the ground?

The rock has a height y at time t according to

y = 25 m + (60 m/s) sin(32°) t - 1/2 gt ²

where g = 9.8 m/s² is the magnitude of the acceleration due to gravity.

Solve for t when y = 0. The quickest way to do this is with the quadratic formula.

0 = 25 m + (60 m/s) sin(32°) t - 1/2 gt ²

t = (-(60 m/s) sin(32°) + √[(60 m/s)² sin²(32°) - 4 (25 m) (-1/2 gt ²)]) / 2

(ignoring the negative root because that would make t negative)

or approximately

t ≈ 7.2 s

b

explanation:

d

explanation:

v=uf/(u-f)

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

=50/-5

=-10 cm (formed at other side)