Question

Students in a physics class are studying how the angle at which a projectile is launched on level ground affects the projectile's hang time and horizontal range. Hang time can be calculated using the formula $t=2v\cdot \sin \left(\theta \right)g$, where t is the hang time in seconds, v is the initial launch velocity, $\theta$ is the projectile angle with respect to level ground, and g is the acceleration due to gravity, defined as $9.8m/s^2$. The horizontal range can be calculated using the formula $R=\frac{v^2\sin \left(2\theta \right)}{g}$, where R is the distance the projectile travels from the launch site, in feet. which of the following gives the value of v, in terms of R, t, and $\theta$?

• A. $v=\frac{t\sin \left(\theta \right)}{2R\sin \left(\theta \right)}$
• B. $v=\frac{2t\sin \left(\theta \right)}{R\sin \left(\theta \right)}$
• C. $v=\frac{2R\sin \left(\theta \right)}{t\sin \left(2\theta \right)}$
• D. $v=\frac{2R\sin \left(2\theta \right)}{t\sin \left(\theta \right)}$

Answer 1

The correct option is C $v=\frac{2R\sin \left(\theta \right)}{t\sin \left(2\theta \right)}$

There is a slight error in the question. It should be

$t=\frac{2vsin\left(\theta \right)}{g}$

And, $R=\frac{v^{2}sin\left(2\theta \right)}{g}$

Now, $\frac{2R}{t}=\frac{{2v}^{2}sin\left(2\theta \right)g}{2gvsin\left(\theta \right)}=\frac{vsin\left(2\theta \right)}{sin\left(\theta \right)}$

So, $v=\frac{2Rsin\left(\theta \right)}{tsin\left(2\theta \right)}$