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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answered: 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]

answered: Guest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

answered: Guest

Ireally don’t know