An object is thrown downward with an initial speed of 5 m/s from a height of 78 m above the ground. At the same instant, a second object is propelled vertically from ground level with a speed of 50 m/s.

The acceleration of gravity is 9.8 m/s2. At what height above the ground will the two objects pass each other?
Answer in units of m.

The height of each is

h1=78+(-5)t-4.9t^2
h2=0+50t-4.9t^2
set them equal, solve for t.
Then put time t into either, to solve for final height.

The acceleration due to gravity on planet X is

one fifth that on the surface of the earth.
If it takes 3.9 s for an object to fall a certain
distance from rest on earth, how long would
it take to fall the same distance on planet X?

To find the height at which the two objects will pass each other, we need to calculate the time it takes for each object to reach that height.

First, let's find the time it takes for the first object (thrown downward) to reach the ground. We can use the equation:

h = 0.5 * g * t^2

Where h is the initial height (78 m), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time. Rearranging the equation to solve for t, we get:

t = sqrt(2 * h / g)

Substituting the given values, we can calculate the time it takes for the first object to reach the ground:

t_1 = sqrt(2 * 78 / 9.8) = 4 s

Next, let's find the time it takes for the second object (propelled upwards) to reach the height at which they will pass each other. The equation for the height reached by the second object can be given by:

h = v_i * t + 0.5 * g * t^2

Where h is the unknown height, v_i is the initial velocity (50 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time. We can rewrite this equation as:

0.5 * g * t^2 + v_i * t - h = 0

This is a quadratic equation, and we can solve it using the quadratic formula:

t_2 = (-v_i + sqrt(v_i^2 + 2 * g * h)) / g

Substituting the given values, we have:

t_2 = (-50 + sqrt(50^2 + 2 * 9.8 * h)) / -9.8

Now, since we know that the two objects will pass each other at the same time, t_1 = t_2. We can set up an equation and solve for h:

t_1 = t_2

4 = (-50 + sqrt(50^2 + 2 * 9.8 * h)) / -9.8

Multiplying both sides by -9.8:

-39.2 = -50 + sqrt(50^2 + 2 * 9.8 * h)

Rearranging the equation:

sqrt(50^2 + 2 * 9.8 * h) = 50 - 39.2

Squaring both sides:

50^2 + 2 * 9.8 * h = (50 - 39.2)^2

Simplifying the equation:

2500 + 2 * 9.8 * h = 10.8^2

2 * 9.8 * h = 116.64 - 2500

2 * 9.8 * h = -2383.36

h = -2383.36 / (2 * 9.8)

h ≈ -122.4 m

Since height cannot be negative, it means that the two objects will not pass each other above the ground. In this scenario, the objects miss each other.