Physics, 26.10.2020 14:00 lilpetals

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

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Answer from: lucilita2017

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

for t = 1 second,

s = 15(1) - 4.9(1)²

s = 15 - 4.5

s = 11.1 m [at t = 1 second]

for t = 4 seconds,

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

for time = 1 second,

v = 15 - 9.8

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

for time = 4 seconds,

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]

Answer from: Quest

because if the tool that you are using is not balanced it there would be false mass.

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge

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You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a...