Question

a ball is thrown at an angle of 45 degrees to the ground, and lands 65 meters away. the initial speed of the ball was m/sec. submit answer

Explanation:

Step1: Recall range formula for projectile motion

The range formula for a projectile launched with initial speed $v_0$ at an angle $\theta$ is $R=\frac{v_0^{2}\sin2\theta}{g}$, where $R$ is the range, $g = 9.8\ m/s^{2}$ is the acceleration due to gravity and $\theta$ is the angle of projection.

Step2: Substitute given values

Given $\theta = 45^{\circ}$, so $\sin2\theta=\sin(2\times45^{\circ})=\sin90^{\circ}=1$, and $R = 65\ m$, $g=9.8\ m/s^{2}$. We have $65=\frac{v_0^{2}\times1}{9.8}$.

Step3: Solve for $v_0$

Cross - multiply to get $v_0^{2}=65\times9.8 = 637$. Then $v_0=\sqrt{637}\approx 25.2\ m/s$.

Answer:

$25.2$