Question

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?

Answer 1

Answer:

(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 2

Answer:

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

$g=-9.8m/s^2$