82.5k views
5 votes
A rock is thrown vertically upward with a speed of 12.0m/s. Exactly 1.00s later, a ball is thrown up vertically along the same path with a speed of 18.0m/s. (a) At what time will they strike each other

User TheJuls
by
4.5k points

2 Answers

5 votes

Final answer:

To find the time when the rock and ball strike each other after being thrown upward with different initial velocities and at different times, we must use kinematic equations to equate their positions and solve for time.

Step-by-step explanation:

The question involves the concept of kinematics in physics, where two objects are thrown upward with different initial velocities and at different times. The goal is to determine at what time they will collide assuming they are on the same vertical path. Since the rock and the ball have different initial velocities and are thrown at different times, we need to set up equations for their motions and find the point in time where their heights are equal.

Strategy

In order to solve for the time at which the rock and the ball strike each other, we can write two equations for their positions (heights) relative to time. The rock is thrown first with an initial velocity v0,rock = 12.0 m/s, and the ball is thrown one second later with v0,ball = 18.0 m/s. We use the equation for vertical motion y = y0 + v0t - (1/2)gt2, where g is the acceleration due to gravity (9.80 m/s2), y0 is the initial height (which is 0 in this case), and t is the time in seconds after the object is thrown.

For the rock: yrock = 12.0t - (1/2)(9.80)t2

For the ball (thrown one second later, so it's t - 1): yball = 18.0(t - 1) - (1/2)(9.80)(t - 1)2

The two objects will strike each other when yrock = yball, which means we need to solve for t when these two equations are equal to each other.

User Ryan Gaudion
by
4.8k points
1 vote

Answer:

Ok, remember that the only force acting on the objects will be the acceleration force.

First, let's describe the motion of the rock.

The acceleration will be:

Ar(t) = -9.8m/s^2

(where the negative sign is because the gravity pulls the rock downwards).

If we want to obtain the velocity of the rock, we should integrate over time and get:

Vr(t) = (-9.8m/s^2)*t + v0

Where v0 is the initial velocity, and we know that it is:

v0 = 12.0m/s

Then:

Vr(t) = (-9.8m/s^)*t + 12m/s.

And the position can be obtained if we integrate again.

Pr(t) = (-4.9m/s^)*t^2 + 12m/s*t + p0

Where p0 is the initial position.

Now let's do the same for the ball, but the ball is thrown one second after, so we will work with the variable (t - 1s)

Ab(t) = -9.8m/s^2

Vb(t) = (-9.8m/s^2)*(t - 1s) + 18m/s

Pb(t) = (-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) + p0

As the ball is thrown along the same path than the rock, the initial position is the same in both cases.

Now the two objects will impact each other when:

Pb(t) = Pr(t)

(-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) + p0 = (-4.9m/s^)*t^2 + 12m/s*t + p0

Let's solve this for t.

(-4.9m/s^2)*(t - 1s)^2 + 18m/s*(t - 1s) = (-4.9m/s^)*t^2 + 12m/s*t

(-4.9m/s^2)*(t^2 -2*t*1s + 1s^2) + 18m/s*(t - 1s) = (-4.9m/s^)*t^2 + 12m/s*t

(-4.9m/s^2)*(-2*t*1s + 1s^2) + 18m/s*t - 18m = 12m/s*t

-4.9m + 9.8m/s*t + 18m/s*t - 18m = 12m/s*t

-22.9m = t*(12m/s - 9.8m/s - 18m/s)

22.9m = t*16.8m/s

(22.9/16.8)s = 1.363 s

So the ball and the rock will strike each other at t = 1.363 seconds.

User Zeddex
by
4.3k points