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## Homework Statement

A person with a 3 kg ball at rest in his hand, throws it with constant force at an angle of 45 degrees. The ball moves in a straight line until it leaves his hand at 2.1 meters above the ground. His arm has the length of 0.55 meters. Between the ball leaving his hand and until impact the total time elapsed is 2.15 seconds and the horizontal distance is 16 meters.

What is the speed of the ball the moment it leaves his hand?

## Homework Equations

Constant acceleration on a straight line:

[itex]v_x=v_{0x}+a_xt[/itex]

[itex]x=x_{0}+v_{0x}t+\frac{1}{2}a_{x}t^2[/itex]

[itex]v^2_x=v^2_{0x}+2a_x(x-x_0)[/itex]

[itex]x-x_0=(\frac{v_{0x}+v_x}{2})t[/itex]

Projectile motion:

[itex]x=(v_0 \cos(\alpha_0))t[/itex]

[itex]y=(v_0 \sin(\alpha_0))t-\frac{1}{2}gt^2[/itex]

[itex]v_x=v_0 \cos(\alpha_0)[/itex]

[itex]v_y=v_0 \sin(\alpha_0)-gt[/itex]

## The Attempt at a Solution

The text tells me that from the ball is at rest and until it leaves his hand, the movement is a straight line. Therefore I'm thinking I should use one of the equation for straight line.

Also, if the force is constant it means the acceleration is constant as well (can somebody please confirm this to me?)

So I want to find [itex]v_x[/itex]. I rearrange my coordinate system so that the straight line movement his arm makes is parallel to the x-axis.

Thus the distance that the ball travels in this straight line is the length of the guy's arm, which is 0.55 meters so:

[itex]x-x_0=0.55 m[/itex]

I also know that since the ball is initially at rest:

[itex]v_{0x}=0[/itex]

But I don't know the acceleration during the straight line motion nor do I know the time it takes for the ball from rest until it leaves his hand. Therefore I have no idea which equation to use.

Maybe I should be looking at the projectile motion equations instead?