A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground. (a) How fast was the ball originally moving when it was kicked. (b) How much longer would it take the ball to reach the ground?

Respuesta :

Answer:

(a)    vo = 24.98m/s

(b)    t = 5.09 s

Explanation:

(a) In order to calculate the the initial speed of the ball, you use the following formula:

[tex]y=y_o+v_ot-\frac{1}{2}gt^2[/tex]      (1)

y: vertical position of the ball = 2.44m

yo: initial vertical position = 0m

vo: initial speed of the ball = ?

g: gravitational acceleration = 9.8m/s²

t: time on which the ball is at 2.44m above the ground = 5.00s

You solve the equation (1) for vo and replace the values of the other parameters:

[tex]v_o=\frac{y-y_o+1/2gt^2}{t}[/tex]        

[tex]v_o=\frac{2.44m-0.00m+1/2(9.8m/s^2)(5.00s)^2}{5.00s}\\\\v_o=24.98\frac{m}{s}[/tex]

The initial speed of the ball is 24.98m/s

(b) To find the time the ball takes to arrive to the ground you use the equation (1) for y = 0m (ground) and solve for t:

[tex]0=24.98t-\frac{1}{2}(9.8)t^2\\\\t=5.09s[/tex]

The time that the ball takes to arrive to the ground is 5.09s

We have that for the Question, it can be said that the speed of  ball and How much longer would it take the ball to reach the ground is

  • u=25.13m/s
  • X=0.095sec

From the question we are told

A goalie kicks a soccer ball straight vertically into the air. It takes 5.00 s for the ball to reach its maximum height and come back down to the level of the crossbar. Assume the crossbar of a soccer goal is 2.44 m above the ground.

(a) How fast was the ball originally moving when it was kicked.

(b) How much longer would it take the ball to reach the ground?

a)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\2.44=ut+1/2(9.8)(5)^2\\\\u=25.13m/s\\\\[/tex]

b)

Generally the Newton equation for the Motion  is mathematically given as

[tex]S=ut+1/2at^2\\\\Therefore\\\\t=\frac{-24}{a}\\\\t=\frac{-2*25.013}{9.81}\\\\t=5.095sec\\\\[/tex]

Therefore

[tex]X=5.095-5[/tex]

X=0.095sec

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