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# 2 dimensional motion problem

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.)

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/s2            x = 42 m

x = 0 m

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

xf = xi + vit + 1/2at2

xf - xi = vit + 1/2at2

Δx = vit + 1/2at2

x/t -at/2 = vi

vi = 0/2.9 - (-9.8)*2.9/2

vi = 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 = 46o

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