A projectile is projected from the origin with a velocity of 30.0 m/s at an angle of 45.0 degrees above the horizontal. What is the velocity and time when the projectile hits the ground? A) V_x = 21.2 m/s, V_y = −21.2 m/s at t = 4.33 s B) V_x = 16.7 m/s, V_y = −16.7 m/s at t = 3.33 s C) V_x = 21.1 m/s, V_y = −16.7 m/s at t = 2.83 s D) V_x = 16.7 m/s, V_y = −21.2 m/s at t = 4.33 s E) V_x = 21.2 m/s, V_y = −15.0 m/s at t = 3.33 s

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Quizplus · Accepted Answer

The velocity components can be calculated using the initial velocity and the angle of projection. For a projectile launched at 45 degrees, the horizontal (Vx) and vertical (Vy) components of the initial velocity are equal. Using $V = 30.0 \, \text{m/s}$ and $\theta = 45^\circ$, we find $Vx = Vy = 30 \cos(45^\circ) = 21.2 \, \text{m/s}$. The time of flight for a projectile launched and landing at the same height, ignoring air resistance, is given by $t = \frac{2V \sin(\theta)}{g}$, where $g = 9.8 \, \text{m/s}^2$. Substituting the given values, $t = \frac{2 \times 30 \sin(45^\circ)}{9.8} \approx 4.33 \, \text{s}$. The horizontal velocity remains constant throughout the flight because there is no horizontal acceleration, so $Vx = 21.2 \, \text{m/s}$. The vertical velocity upon hitting the ground will be equal in magnitude but opposite in direction to its initial vertical velocity, so $Vy = -21.2 \, \text{m/s}$.

A Projectile Is Projected from the Origin with a Velocity