# A ball is thrown such that its velocity has an upward component of 0.5 m/s and horizontal component of 3.6 m/s. What is the ball's speed and at what angle was it thrown?

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Let the magnitude of the velocity at which the ball is thrown be V and the angle with the horizontal it is thrown at equal X.

The vertical component of the velocity of the ball is given by V*sin X and the horizontal component of the velocity of of the ball is V*cos X.

As the vertical component of the ball's velocity is given as 0.5 m/s and the horizontal component is given as 3.6 m/s, the magnitude of the velocity it is throw at is sqrt(0.5^2 + 3.6^2) = sqrt(0.25 + 12.96) = sqrt(13.21) m/s

The angle at which the ball is thrown is equal to an angle of arc tan (0.5/3.6) = 7.907 degrees with the horizontal.

The vertical component of the ball velocity = 0.5 m/s

The Horizontal velocity of the ball velocity = 3.6 m/s

*The resultant speed of the ball **is equal **to the vector sum of the two components and is equal to sqrt(0.5^2+3.6^2) m/s is equal to*

*3.63 m/s***The direction of the ball with the horizontal **is an angle given by arc tan(vertical component / horizontal component or arc tan(0.5/3.6) and** is equal to 7.91 degrees**