Two stones are thrown vertically upward from the ground, one with three times the initial speed of the other. Assume free fall.

If the faster stone takes 12.0s to return to the ground, how long will it take the slower stone to return? i got the answer it 4s


If the slower stone reaches a maximum height of H, how high (in terms of H) will the faster stone go?

Respuesta :

we are given the data two stones thrown at different velocities results to different times. We can use the equation y = vot - 1/2 gt2. At maximum height, dy/dt = 0 = vo - gt. If t2 is 12 s , then vo is equal to 117.6 m/s. Then, the time of t1 is equal to 117.6*3/9.8 equal to 36 seconds. When t = 36 seconds, then y = 117.6*36 - 1/2 *9.8*36^2 equal to -2116.8 m  

It will take 4 seconds for the slower stone to return to the ground.

The faster stone will reach a maximum height of 9 H

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 !

Given:

v₁ = 3 v₂

t₁ = 12 s

h₂ = H

Unknown:

t₂ = ?

h₁ = ?

Solution:

We can use the following formula to calculate time taken to reach the ground,

[tex]h = v ~ t - \frac{1}{2} ~ g ~ t^2[/tex]

[tex]0 = v ~ t - \frac{1}{2} ~ g ~ t^2[/tex]

[tex]v ~ t = \frac{1}{2} ~ g ~ t^2[/tex]

[tex]v = \frac{1}{2} ~ g ~ t[/tex]

[tex]2 ~ v = g ~ t[/tex]

[tex]\large {\boxed {t = \frac{2 ~ v}{g} } }[/tex]

Next, we will compare the time taken by these two stones to reach the ground

[tex]\frac{t_1}{t_2} = \frac{2 ~ v_1 / g}{2 ~ v_2 / g}[/tex]

[tex]\frac{t_1}{t_2} = \frac{v_1}{v_2}[/tex]

[tex]\frac{12}{t_2} = \frac{3 v_2}{v_2}[/tex]

[tex]\frac{12}{t_2} = 3[/tex]

[tex]t_2 = 12 \div 3[/tex]

[tex]\boxed {t_2 = 4 ~ seconds}[/tex]

We can use the following formula to find the maximum height reached by these two stones

[tex]v_f ~ ^2 = v ~ ^2 - 2 g h[/tex]

[tex]0^2 = v ~ ^2 - 2 g h[/tex]

[tex]2 g h = v ~ ^2[/tex]

[tex]\large {\boxed {h = \frac{v ~ ^2}{2 g} } }[/tex]

Next, we will compare the maximum height reached by these two stones

[tex]\frac{h_1}{h_2} = \frac{v_1 ~ ^2 / 2 g}{v_2 ~ ^2 / 2 g}[/tex]

[tex]\frac{h_1}{h_2} = \frac{v_1 ~ ^2}{v_2 ~ ^2}[/tex]

[tex]\frac{h_1}{H} = \frac{(3 v_2) ~ ^2}{v_2 ~ ^2}[/tex]

[tex]\frac{h_1}{H} = \frac{9 v_2 ~ ^2}{v_2 ~ ^2}[/tex]

[tex]\frac{h_1}{H} = 9[/tex]

[tex]\boxed {h_1 = 9 H}[/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

Ver imagen johanrusli