Physics, 05.10.2019 06:00 nicole667

You throw a ball vertically upward so that it leaves the ground with velocity +5.00 m/s. (a) what is its velocity when it reaches its maximum altitude? (b) what is its acceleration at this point? (c) what is the velocity with which it returns to ground level? (d) what is its acceleration at this point?

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Answer from: maddieeelllis3956

(a) The velocity (v) of the ball when reaches its maximum altitude is zero .

v= 0

(b) The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball is g =-9,8 m/s².

(c)velocity of the ball when it returns to the ground :

$v_{f} =5 \frac{m}{s}$ in direction -y

$v_{f} =-5 \frac{m}{s}$

(d)a=g = -9,8 m/s² : acceleration of the ball when it returns to the ground

Explanation:

Ball Kinematics

We apply the free fall formula:

vf²=v₀²+2*a*y Formula (1)

y:displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration

Data

v₀ = +5.00 m/s

Problem development

(a) What is its velocity when it reaches its maximum altitude?

in ymax, vf=0 , At maximum height the velocity is zero and the ball falls freely

(b) What is its acceleration at this point?

The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball is 9,8 m/s².

g= -9.8 m/s²

c) What is the velocity with which it returns to ground level?

We apply the Formula (1) to calculate the maximum height (h):

Ball movement up

vf²=v₀²+2*a*h

0 = (5)² +2*(-9.8)*h

19.6*h = 25

h= 25/19.6 = 1.275 m

Ball movement down

The distance the ball goes up is equal to the distance it goes down

h= 1.275 m , v₀=0

vf²=v₀²+2*a*h

vf²=0+2*(-9.8)*(-1.275)

$v_{f} = \sqrt{25}$

$v_{f} =5 \frac{m}{s}$ in direction -y

$v_{f} =-5 \frac{m}{s}$

(d) What is its acceleration at this point?

a is constant , a= g = -9.8 m/s²

Answer from: crayons18

a) $v_{y}=0$

b) a=g=-9.8m/s^2

c) $v_{y}=-5m/s$

d) $a_{y}=g=-9.8m/s^2$

Explanation:

According to the exercise the ball is thrown upward at an initial velocity of 5m/s

a) At maximum altitude the ball stops going up and then start going down

That means that the velocity at that point is 0

$v_{y}=0$

b) The acceleration that takes place here is gravity. And it's at the y-axis

a=g=-9.8m/s^2

c) To calculate the velocity which it returns to the ground we need to calculate how much time does it take to do it first

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

If the ball is on ground level y=0

$0=0+5t-\frac{1}{2}(9.8)t^2$

Using quadratic equation we solve for t

t=0 or t=1.02s

Since time can not be 0 the answer is t=1.02s. Now, we can calculate the velocity which it returns to the ground

$v_{y}=v_{oy}+gt$

$v_{y}=5m/s-(9.8m/s^2)(1.02s)=-5m/s$

d) According to free falling objects the acceleration at every moment for the fall of an objet is gravity

g=-9.8m/s^2

Answer from: Quest

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

Look at the pendulum diagram between which points does the pendulum’s potential energy increase most

Answer from: Quest

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.

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You throw a ball vertically upward so that it leaves the ground with velocity +5.00 m/s. (a) what is...

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