How do you calculate the horizontal velocity of an object in a projectile problem?

it depends on what you have given. If you know a projectile is fired with velocity v at an angle theta, then horizontal velocity is

Vh=VcosTheta

It's weird because I'm only give 25m and 5m in this problem. It looks like this kinda

|
| 5m
|
|_________________
25m

I'm suppose to find the horizontal velocity with this

You dropped it from 5 meters

and it went 25 meters maybe ??
If so:
time to drop 5 meters
5 = (1/2)(9.8)t^2
t = 1.43 seconds
u = 25 m/1.43 s = 17.5 m/s

To calculate the horizontal velocity of an object in a projectile problem, you need to have information about the initial velocity and the launch angle. The horizontal velocity, also known as the x-component of velocity, remains constant throughout the motion.

Here's the step-by-step procedure to calculate the horizontal velocity:

1. Identify the given values: You need to know the initial velocity (also called the launch or initial speed) of the object and the launch angle at which it is projected. Let's say the initial velocity is represented as 'V0' and the launch angle as 'θ'.

2. Resolve the initial velocity: The initial velocity can be resolved into its horizontal (Vx) and vertical (Vy) components. Since we want to find the horizontal velocity, we only consider the horizontal component. The horizontal component is given by Vx = V0 * cos(θ), where 'cos' represents the cosine function.

3. Calculate the horizontal velocity: Substitute the values of V0 and θ into the formula from the previous step. Use a calculator to find the cosine value of the launch angle since it is likely in degrees. Multiply the initial velocity by cos(θ) to get the horizontal velocity (Vx).

For example, let's say the initial velocity is 30 m/s and the launch angle is 45 degrees. Plugging these values into the equation Vx = V0 * cos(θ), we get:

Vx = 30 m/s * cos(45°)
Vx = 30 m/s * 0.7071 ≈ 21.2 m/s

So, the horizontal velocity of the object is approximately 21.2 m/s. Remember, this value will remain constant throughout the projectile motion.