Question

2. An archer shoots an arrow at 83.0 m/s at a 62.0 degree angle. If the ground is flat, how much time is the arrow in the air?

Answer 1

Answer:

t=14.96 sec

Explanation:

Diagonal Launch

It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.

The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is :

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the initial speed, $\theta$ is the angle, t is the time and g is the acceleration of gravity .

In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus $y_o=0$, and:

$\displaystyle y=v_osin\theta \ t-\frac{gt^2}{2}$

The value of y is zero twice: when t=0 (at launching time) and in t=$t_f$ when it goes back to the ground. We need to find that time $t_f$ by making $y=0$

$\displaystyle 0=v_osin\theta\ t_f-\frac{gt_f^2}{2}$

Dividing by $t_f$

$\displaystyle v_osin\theta=\frac{gt_f}{2}$

Then we find the total flight time as

$\displaystyle t_f=\frac{2v_osin\theta}{g}$

$\displaystyle t_f=\frac{2(83)sin\ 62^o}{9.8}$

$\displaystyle t_f=14.96\ sec$