Question

at what time do two objects one falling from a height of 15 m and the other launched upward from a ground in rising to a height of 60 m cross path?

Answer 1

To determine when two objects, one falling from a height of 15 m and the other launched upward from the ground to a height of 60 m, cross paths, we can analyze their respective motions and find the time at which their heights are equal.

Step-by-step explanation:

In order to determine when the two objects cross paths, we need to analyze their respective motions and find the time at which their heights are equal.

For the object falling from a height of 15 m, we can use the equation h = 1/2gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.

For the object launched upward from the ground, we can use the same equation, but with negative values for h and g, since it is moving in the opposite direction. By setting the two equations equal to each other and solving for t, we can find the time at which the heights are equal.

Answer 2

To determine at what time two objects, one falling from a height of 15 m and the other launched upward from the ground to a height of 60 m, cross paths, we can use the concept of projectile motion. The object falling takes about 1.77 s to hit the ground, and the object launched upward takes about 3.19 s to reach the height of 60 m. Therefore, the two objects will cross paths at approximately 1.77 s after the object falling is released.

Step-by-step explanation:

To determine at what time two objects, one falling from a height of 15 m and the other launched upward from the ground to a height of 60 m, cross paths, we can use the concept of projectile motion. When an object is launched upward or dropped from a height, it follows a parabolic path. The time at which the two objects cross paths can be determined by comparing their vertical positions at different times.

Let's assume that the object falling from a height of 15 m takes time 't' to hit the ground. We can use the equation h = ut + (1/2)gt^2, where h is the vertical displacement, u is the initial velocity, g is the acceleration due to gravity, and t is the time. For the object falling, h = 15 m, u = 0 m/s, and g = 9.8 m/s^2. Solving the equation for t, we get t = √(2h/g) = √(2*15/9.8) ≈ 1.77 s.

Now, for the object launched upward, let's assume it takes time 'T' to reach a height of 60 m. Since it is launched upward, its initial velocity, u = 0 m/s, and g = -9.8 m/s^2 (negative due to the opposite direction of motion). Using the same equation as before, h = 60 m, u = 0 m/s, and g = -9.8 m/s^2, we can solve for time 'T'. We get T = √(2h/g) = √(2*60/-9.8) ≈ 3.19 s.

Therefore, the two objects will cross paths at approximately 1.77 s after the object falling from a height of 15 m is released.