A ball launched from ground level lands 2.9 s later on a level field 42 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)

## 2 dimensional motion problem

# 1 Answer

Begin by separating the known information into horizontal and vertical components.

Vertical: Horizontal:

t = 2.9 s t = 2.9 s

a = -9.81 m/s^{2} x = 42 m

x = 0 m

Looking only at the vertical information, we can determine the initial velocity (vertically) using the formula:

x_{f} = x_{i} + v_{i}t + 1/2at^{2}

x_{f} - x_{i} = v_{i}t + 1/2at^{2}

Δx = v_{i}t + 1/2at^{2}

x/t -at/2 = v_{i}

v_{i} = 0/2.9 - (-9.8)*2.9/2

v_{i} = 14.2 m/s

That is the vertical vector, next we are going to look for the horizontal vector.

Since there is no
**horizontal** acceleration on the ball after it is launched, we will look for the average velocity.

v = x/t

v = 42/2.9

v = 14.5 m/s

To find the resultant vector, use the Pythagorean Theorem:

v = √((14.2 m/s)^{2} + (14.5 m/s)^{2})

v
= 20. ms

*Rounded
to 2 significant figures in the answer.

To
find the angle of the velocity, we will use the definition of the tangent. Since the opposite is the vertical component and the adjacent is the horizontal component:

Tanθ
= 14.5/14.2

Tanθ
= 1.02

θ
= 46^{o}

The
projectile was fired with an initial velocity of 20. m/s at 46^{o}.

## Comments

Sorry if the answer could use more explanation. I had to take some of my explanation because of the 10,000 character limit.

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