A projectile is projected from the origin with a velocity of $25.0 \mathrm{~m} / \mathrm{s}$ at an angle of 45.0 degrees above the horizontal. What is the time it takes for the projectile to hit the ground?
A) $2.22 \mathrm{~s}$
B) $1.95 \mathrm{~s}$
C) $3.04 \mathrm{~s}$
D) $3.61 \mathrm{~s}$
E) $2.75 \mathrm{~s}$

Question

Quizplus · Accepted Answer

The time of flight for a projectile launched at an angle $\theta$ with initial velocity $v_0$ is given by $T = \frac{2v_0 \sin \theta}{g}$. Substituting $v_0 = 25.0 \, \mathrm{m/s}$, $\theta = 45^\circ$, and $g = 9.81 \, \mathrm{m/s^2}$, we find $T \approx 3.61 \, \mathrm{s}$.

A Projectile Is Projected from the Origin with a Velocity 25.0 m/s25.0 \mathrm{~m} / \mathrm{s}25.0 m/s

A Projectile Is Projected from the Origin with a Velocity $25.0 \mathrm{~m} / \mathrm{s}$