Apair of glasses is dropped from the top of a 42.0 meter high stadium. a pen is dropped from the same position 2.00 seconds later. how high above the ground is the pen when the glasses hit the ground? (disregard air resistance. a = - g = - 9.81 m/s ^2)

Given that,

For glasses:

distance, s = 42 m

a= -g = -9.81 m/s²

we can calculate time taken by glasses to hit the ground as follow:

s= Vi* t + 1/2 at²

Since at top initial velocity, Vi= 0

42 = 0 - 0.5* 9.81 t²

t = 2.926 s

Pen is drop 2 second after the glasses is dropped. So time difference will be:

Δt = 2.926 - 2 = 0.926 s

Distance covered by Pen in 0.926 s can be calculated as:

s = 0 - 0.5* 9.81 * 0.926²

s= 4.2 m

Distance of pen from ground will be, 42-4.2 = 37.8 m

When the glasses hit the ground, the pen will be 37.8 m above the ground.

d

explanation:

v=uf/(u-f)

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

=50/-5

=-10 cm (formed at other side)

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