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?

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

Acceleration is rate of change of velocity.

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,

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

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

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

Grade: High School

Subject: Physics

Chapter: Kinematics

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

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

goog may be able to you find that answer