a target t lies flat on the ground 3m from the side of a building that is 10m tall, as shown above. a student rolls a ball off the horizontal speed with which the ball must leave the roof if it is to strike the target moest nearly?

how long does it take to fall 10m?

Then, using that value of t, solve for v in

vt = 3

since the ball must travel horizontally for 3m in that amount of time.

1.41

To find the horizontal speed with which the ball must leave the roof, we can use the concept of projectile motion. The ball will travel horizontally (parallel to the ground) after leaving the roof, and its vertical motion will be affected by gravity.

Let's break down the problem step-by-step:

Step 1: Determine the time it takes for the ball to reach the ground.
To find the time, we can use the equation for vertical motion:
y = yo + vot + 0.5at^2

In this case, yo (initial vertical position) is 10m (height of the building), and y (final vertical position) is 0m (ground level). The acceleration due to gravity, a, is approximately -9.8 m/s^2 (negative because it acts downward). We can neglect the initial vertical velocity (vo) because the ball is initially at rest (lying flat on the roof). So the equation becomes:
0 = 10 + 0 - 4.9t^2

Simplifying the equation:
4.9t^2 = 10

Solving for t:
t^2 = 10 / 4.9
t^2 ≈ 2.04
t ≈ √2.04
t ≈ 1.43 s

Step 2: Calculate the horizontal distance traveled by the ball.
The horizontal distance traveled by an object in projectile motion is given by:
x = vxt
where vx is the horizontal component of velocity and t is the time.

In this case, x (horizontal distance) is 3m, and t (time) is 1.43s.

Solving for vx:
vx = x / t
vx = 3 / 1.43
vx ≈ 2.10 m/s

Therefore, the horizontal speed at which the ball must leave the roof to strike the target most nearly is approximately 2.10 m/s.

To find the horizontal speed with which the ball must leave the roof in order to strike the target most nearly, we need to analyze the motion of the ball.

First, let's break down the problem:

1. The ball is rolling off a roof, which means it will follow a projectile motion in two dimensions - horizontal and vertical.
2. The horizontal motion is independent of the vertical motion due to the absence of any horizontal forces acting on the ball.
3. The vertical motion of the ball is affected by gravity.

Now, let's find the time it takes for the ball to hit the target. Since the ball travels horizontally, the time of flight will only depend on the horizontal distance and the horizontal speed.

The horizontal distance between the target and the building is 3 meters. Therefore, the time of flight can be found using the formula:

time = distance / horizontal speed

We can rearrange the formula as follows:

horizontal speed = distance / time

Since the time of flight for a projectile depends only on the vertical motion, we need to analyze the vertical component of the motion to find the time of flight.

We know that the ball leaves the roof of a 10-meter building. The vertical distance traveled by the ball is equal to the height of the building. The formula to calculate this distance considering only gravitational force is:

height = (1/2) * g * (time)^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

We can rearrange the formula to solve for time:

time = sqrt(2 * height / g)

Substitute this value into the initial formula for horizontal speed:

horizontal speed = distance / sqrt(2 * height / g)

Substituting the given values:

distance = 3 meters
height = 10 meters
g = 9.8 m/s^2

horizontal speed = 3 / sqrt(2 * 10 / 9.8)

Simplifying further:

horizontal speed = 3 / sqrt(20/9.8)
horizontal speed = 3 / sqrt(2.0408)
horizontal speed ≈ 3 / 1.43
horizontal speed ≈ 2.10 m/s

Therefore, the ball must leave the roof with a horizontal speed of approximately 2.10 m/s to strike the target most nearly.