A projectile is projected from the origin with a velocity of $30.0 \mathrm{~m} / \mathrm{s}$ at an angle of 40.0 degrees above the horizontal. What is the time it takes the projectile to reach maximum height?
A) $1.02 \mathrm{~s}$
B) $1.97 \mathrm{~s}$
C) $1.54 \mathrm{~s}$
D) $2.23 \mathrm{~s}$
E) $2.66 \mathrm{~s}$

Question

Quizplus · Accepted Answer

The time to reach maximum height can be calculated using the formula $t = \frac{v_0 \sin \theta}{g}$, where $v_0$ is the initial velocity, $\theta$ is the launch angle, and $g$ is the acceleration due to gravity (approximately $9.8 \mathrm{~m/s^2}$). Plugging in the given values: $t = \frac{30.0 \sin(40.0^\circ)}{9.8} \approx 1.97 \mathrm{~s}$.

A Projectile Is Projected from the Origin with a Velocity 30.0 m/s30.0 \mathrm{~m} / \mathrm{s}30.0 m/s

A Projectile Is Projected from the Origin with a Velocity $30.0 \mathrm{~m} / \mathrm{s}$