# A tennis player hits a ball at ground level with a velocity of 46.5 m/s and the range is 213.11 meters. What is the launch angle?

justaguide | Certified Educator

The tennis player hits the ball at ground level with a velocity of 46.5 m/s. Let the launch angle be A. The velocity at which the ball is hit can be divided into two components, a horizontal component given by 46.5*cos A and a vertical component equal to 46.5*sin A.

There is an acceleration due to gravity equal to 9.8 m/s^2 acting in a vertically downwards direction. As the ball rises, the vertical component of its velocity decreases, reaches 0, and then becomes negative. When the ball reaches ground level the vertical component of the velocity is -46.5*sin A.

Use the formula V = U + a*t to determine the time of motion of the ball. V = -46.5*sin A, U = 46.5*sin A and a = -9.8 m/s^2

-46.5*sin A = 46.5*sin A - 9.8*t

=> t = `(93*sin A)/9.8`

The horizontal range of the ball is given by `((93*sin A)/9.8)*(46.5*cos A)` . This is equal to 213.11 m

=> `213.11 = (46.5*46.5*sin 2A)/9.8`

=> `sin 2A = 0.96588`

=> `2A = sin^(-1)(0.96588)`

=> `2A = 74.989`

=> `A = 37.49`

The launch angle of the ball was 37.49 degrees.

justaguide | Certified Educator

The range of a projectile is given by `R = (v^2*sin 2A)/g` where A is the launch angle and v is the initial velocity.

The range of any projectile is the same for two angles of launch A and A' as `sin (180 - x) = sin x`

From this `sin 2A = sin (2*(90 - A')`

`sin (2*A)` and `sin (2*A')` have the same value if `A + A' = 90`