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.

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

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