You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15 m/s; the ball is then in free fall. on its way back down, it is just misses the railing. At the location of the building, (g=9.8m/s2). Find a) the position and velocity of the ball 1.00 s and 4.00 s after leaving your hand? b) the velocity when the ball is 5.00 m above the railing; c) the maximum height reached and the time at which it is reached; and d) the acceleration of the ball when it is at its maximum height

Analysing the question:

We are given:

initial speed of the ball (u) = 15 m/s

acceleration due to gravity (a) = 9.8 m/s²

Position and the velocity of the ball 1 and 4 seconds after throwing:

Position of the Ball at the respective time:

from the second equation of motion:

s = ut + 1/2at²

replacing the variables

s = 15(t) + 1/2(-9.8)(t)²

s = 15t - 4.5(t)²

        s = 15(1) - 4.9(1)²

        s = 15 - 4.5

        s = 11.1 m  [at t = 1 second]

        s = 15(4) - 4.9(4)²

        s = 60 - 4.9(16)

        s = 60 - 78.4

        s = -18.4 m  [at t = 4 seconds]

Velocity of the ball at the respective time:

from the first equation of motion:

v = u + at

replacing the variables

v = 15 + (-9.8)t

        v = 15 - 9.8

        v = 5.2 m/s   [for t = 1 second]

        v = 15 - (9.8)(4)

        v = 15 - 39.2

        v = -24.2 m/s   [for t = 4 seconds]

Velocity when ball is 5m above the railing:

since the ball is 5m above the railing, s = 5m

from the third equation of motion:

v² - u² = 2as

replacing the variables

v² - (15)² = 2(-9.8)(5)

v² - 225 = -98

v² = 225 - 98

v² = 127

v = 11 m/s (approx)

Maximum height reached and the time at which it was reached:

at the maximum height, the ball will come to rest for a moment. at that moment, the velocity of the ball will be equal to 0 m/s

Maximum height reached:

from the third equation of motion:

v² - u² = 2as

replacing the variables

(0)² - (15)²  = 2(-9.8)(s)        [here, v = 0 m/s since the ball comes to rest at              maximum height]

-225 = -19.6(s)

s = -225 / -19.6

s = 11.5 m (approx)

Time when it was reached:

from the first equation of motion:

v = u + at

replacing the variables

0 = 15 + (-9.8)(t)

-9.8t = -15

t = 15/9.8

t = 1.53 seconds

Acceleration of the ball at maximum height:

at maximum or any other height, a free-falling body has acceleration equal to the acceleration due to gravity [except when an acceleration is applied on the ball]

So, since there was no acceleration applied on the ball. It has an acceleration of -9.8 m/s ['-' for downward direction]

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.