Astone was dropped off a cliff and hit the ground with a speed of 104 ft/s. what is the height of the cliff? (use 32 ft/s2 for the acceleration due to gravity.)

Let the stone is dropped off a cliff of height S m ft.

Given, initial velocity of the stone, u= 0 ft/s

Final velocity of the stone, v= 104 ft/s

Acceleration of the stone= Acceleration due to gravity = 32 ft/s²

Using the third equation of motion,

v² - u² = 2aS

Substituting the values we get,

S= 169 ft

Height of the cliff is 169 ft.

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

3.2x10^7