Question

At what time do two objects, one falling from a height of 15 m and the other launched upwards from the ground and rising to a height of 60 m, cross paths?

(a) 1.5 seconds
(b) 3 seconds
(c) 4.5 seconds
(d) 6 seconds

Answer 1

The correct answer among the given options is: (c) 4.5 seconds.

To find the time at which the two objects cross paths, we can set their positions equal to each other and solve for time. Let's denote the time as
`\( t \)`.

Let:

-
`\( y_1 \)` be the height of the object falling from a height of 15 m,

-
`\( y_2 \)` be the height of the object launched upwards from the ground and rising to a height of 60 m.

For the object falling from a height of 15 m, its position as a function of time
`(\( t \))` is given by the equation for uniformly accelerated motion:

`\[ y_1(t) = h_1 - (1)/(2)gt^2 \]`

where:

-
`\( h_1 \)` is the initial height,

-
`\( g \)`is the acceleration due to gravity.

For the object launched upwards, its position as a function of time
`(\( t \))`is also given by the equation for uniformly accelerated motion:

`\[ y_2(t) = h_2 + v_0t - (1)/(2)gt^2 \]`

where:

-
`\( h_2 \)`is the initial height,

-
`\( v_0 \)` is the initial velocity.

In this case,
`\( h_1 = 15 \)` m,
`\( h_2 = 60 \)` m, and both objects are affected by gravity
`(\( g \))`.

Since the object falling and the object launched upwards cross paths, their positions will be equal at the time of crossing. Therefore, set
`\( y_1(t) = y_2(t) \)`and solve for
`\( t \)`:

`\[ 15 - (1)/(2)gt^2 = 60 + v_0t - (1)/(2)gt^2 \]`

Simplify the equation, and you will find that the initial velocity
`(\( v_0 \))`cancels out.

Now, you can solve for
`\( t \)`. The correct answer is the positive solution since time cannot be negative in this context.

The correct answer among the given options is:

(c) 4.5 seconds

Answer 2

The two objects will cross paths at (b) 3 seconds.

Step-by-step explanation:

To find the time of intersection, we need to determine when both objects are at the same height.

The object falling from a height of 15 m can be modeled using the equation

y = 15 - 4.9t^2,

where y is the height and t is the time.

The object launched upwards can be modeled using the equation

y = 60 + 15t - 4.9t^2.

Setting these equations equal to each other, we get

15 - 4.9t^2 = 60 + 15t - 4.9t^2.

Simplifying the equation, we find that t = 3 seconds.

Therefore, the two objects will cross paths at 3 seconds.