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## Homework Statement

A ball launched from ground level lands 2.85s later on a level field 48 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.)

## Homework Equations

##x=x_0+v_{0x}t+\frac{1}{2}at^2##

##y=y_0+v_{0y}t+\frac{1}{2}at^2##

## The Attempt at a Solution

So, I worked out this problem seemingly without an issue, but WebAssign keeps telling me the answers are wrong, so maybe a bit of new eyes can point out what I've done wrong.

I start by attempting to solve for the x component of the velocity vector:

##x=x_0+v_{0x}t+\frac{1}{2}at^2##

##48m=0m+v_{0x}(2.85s)+\frac{1}{2}(0)(2.85s)^2##

##48m=v_{0x}(2.85s)##

##v_{0x}=\frac{48m}{2.85s}##

##v_{0x}=16.84\frac{m}{s}##

I then found the y component of the velocity:##48m=0m+v_{0x}(2.85s)+\frac{1}{2}(0)(2.85s)^2##

##48m=v_{0x}(2.85s)##

##v_{0x}=\frac{48m}{2.85s}##

##v_{0x}=16.84\frac{m}{s}##

##y=y_0+v_{0y}t+\frac{1}{2}at^2##

##0m=0m+v_{0y}(2.85s)+\frac{1}{2}(-9.8\frac{m}{s^2})(2.85s)^2##

##0=v_{0y}(2.85s)-13.96m##

##v_{0y}=\frac{13.96m}{2.85s}##

##v_{0y}=4.9\frac{m}{s}##

Now that I have both components, I use the Pythagorean Theorem to find the magnitude of the velocity:##0m=0m+v_{0y}(2.85s)+\frac{1}{2}(-9.8\frac{m}{s^2})(2.85s)^2##

##0=v_{0y}(2.85s)-13.96m##

##v_{0y}=\frac{13.96m}{2.85s}##

##v_{0y}=4.9\frac{m}{s}##

##||\vec{v_0}|| = \sqrt{v_{0x}^2+v_{0y}^2}##

##||\vec{v_0}||= \sqrt{(16.84\frac{m}{s})^2+(4.9\frac{m}{s})^2}##

##||\vec{v_0}||=17.5\frac{m}{s}##

Finally, I use these to find the angle at which the ball was fired:##||\vec{v_0}||= \sqrt{(16.84\frac{m}{s})^2+(4.9\frac{m}{s})^2}##

##||\vec{v_0}||=17.5\frac{m}{s}##

##sinx=\frac{4.9}{17.5}##

##x=sin^{-1}(\frac{4.9\frac{m}{s}}{17.5\frac{m}{s}})##

##x=16.22^{\circ}##

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