A pair 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 lat
er. How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance. a = - g = - 9.81 m/s ^2)
1 answer:
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.
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(Answered by Benjemin)
Twice the distance to double an impact
F=ma
F=QE = 1.602e-19C*700N/C = 1.1214e-16N
1.1214e-16N = ma = 1.6726e-27kg * a
a = 6.702e10 m/s² along the direction of the field line