Aprojectile is launched at a 30° angle relative to the ground. the projectile has an initial velocity of 15 m/s and travels through the air for 2 seconds.

the horizontal displacement of the projectile, rounded to the nearest hundredth, is m.

The answer would be:

25.98m

Horizontal distance means you're trying to find x-component in terms of distance or dx. The formula is:

Where:

dx = horizontal distance

Vix = initial horizontal velocity

t = time

Now because the object was thrown at an angle, then that means you need to find the x-component of the given velocity, which you need to solve for the distance. This can be solved with the equation:

Where:

Vi = initial velocity

θ = angle it was thrown

So let's get started with your problem:

Given:

Vi = 15m/s

θ = 30°

t = 2 seconds

Solution:

Now that we've got the horizontal initial velocity, we insert it in the equation for horizontal distance:

Answer : d = 25.98 m

Explanation:

It is given that,

Initial velocity of projectile is,

Angle of projectile,

We have to find the horizontal displacement of the projectile.

Since, velocity is a vector quantity. It can be resolved in two components i.e.

 ( horizontal )

  ( vertical )

velocity,

So,

Hence, horizontal displacement of the projectile is 25.98 m.

i think it's 35 meters