# A cannon ball is launched (from the ground) at an angle of 35 degrees, and lands 2500 m from where it was launched. At what speed was the cannon ball launched?

Apr 5, 2016

${v}_{i} = 161 , 55 \frac{m}{s}$

#### Explanation:

${v}_{i} : \text{initial velocity}$
$\alpha : \text{ angle}$
$g = 9 , 81 \frac{m}{s} ^ 2$
${x}_{m} = 2500 m$

${x}_{m} = \frac{{v}_{i}^{2} \cdot \sin 2 \alpha}{g}$

$2500 = \frac{{v}_{i}^{2} \cdot \sin 2 \cdot 35}{9 , 81}$

${v}_{i}^{2} = \frac{2500 \cdot 9 , 81}{\sin} 70$

v_i^2=24525/(0,9396926208

${v}_{i}^{2} = 26098 , 96$

${v}_{i} = 161 , 55 \frac{m}{s}$