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A ball is thrown vertically upward with a speed of 1.53 m/s from a point 4.21 m above the ground. calculate the time in which the ball will reach the ground.

Respuesta :

skyluke89
The ball moves by uniformly accelerated motion, and its vertical position at time t is described by the following law
[tex]y(t) = h+v_0t - \frac{1}{2}gt^2 [/tex]
where
[tex]h=4.21 m[/tex] is the initial height from which the ball starts its motion
[tex]v_0=1.53 m/s[/tex] is the initial velocity of the ball
[tex]g=9.81 m/s^2[/tex] is the gravitational acceleration

The time in which the ball reaches the ground is the time t at which the vertical position y(t) becomes zero:
[tex]0= h + v_0 t - \frac{1}{2}gt^2[/tex]
Which means
[tex]0=4.21 + 1.53 t - 4.9 t^2 [/tex]

whose solutions are:
[tex]t=-0.78 s[/tex]
[tex]t=1.10 s[/tex]
Neglecting the negative solution (since it has no physical meaning), we can say that the ball reaches the ground after 1.10 s.
johanrusli

The time in which the ball will reach the ground is about 1.10 s

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

This problem is about Kinematics.

We will solve it in the following way

Given:

initial speed = u = 1.53 m/s

initial height = H = 4.21 m

Unknown:

time taken = t = ?

Solution:

[tex]H = ut - \frac{1}{2}gt^2[/tex]

[tex]-4.21 = 1.53t - \frac{1}{2}(9.8)t^2[/tex]

[tex]-4.21 = 1.53t - 4.9t^2[/tex]

[tex]4.9t^2 - 1.53t - 4.21 = 0[/tex]

We will solve the above equation using the following quadratic function formula:

[tex]t = \frac{1.53 + \sqrt{1.53^2 - 4(4.9)(-4.21)}}{2(4.9)}[/tex]

[tex]t \approx 1.10 ~ s[/tex]

Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

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