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madreJ [45]
3 years ago
5

Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be

low, 3.0 seconds later. Projectile B is launched horizontally from the same location at a speed of 30. Meters per second. Approximately how high is the cliff?
Physics
1 answer:
miv72 [106K]3 years ago
4 0

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

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Answer:

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a=\dfrac{(3\times 10^7)^2-0}{2\times 0.02}

a=2.25\times 10^{16}\ m/s^2

According to Newton's law, force acting on the electron is given by :

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F = q E, E = electric field

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I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.

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