The velocity at the midway point of a ball able to reach a height y when thrown with velocity vi at the origin is:
A) $\frac{v_{i}}{2}$
B) $\sqrt{v_{i}^{2} 2 g y}$
C) $\sqrt{\frac{v_{i}^{2}}{2}}$
D) $\sqrt{v_{i}^{2}+2 g y}$
E)gy

Question

Quizplus · Accepted Answer

The velocity at the midway point of a projectile's trajectory can be found using the formula: 
$$v = \sqrt{v_i^2 + 2gy}$$ 
where $v_i$ is the initial velocity, $g$ is the acceleration due to gravity, and $y$ is the maximum height reached by the projectile. 
Since the question gives us the value of $y$ as $y$, we can substitute this value into the formula to get: 
$$v = \sqrt{v_i^2 + 2g(y)}$$ 
At the midway point, the projectile has reached half of its maximum height, so the height is $\frac{1}{2}y$. Substituting this into the formula, we get: 
$$v = \sqrt{v_i^2 + 2g(\frac{1}{2}y)}$$ 
Simplifying, we get: 
$$v = \sqrt{v_i^2 + gy}$$ 
which matches choice C.

The Velocity at the Midway Point of a Ball Able vi2\frac{v_{i}}{2}2vi​​

The Velocity at the Midway Point of a Ball Able $\frac{v_{i}}{2}$