Question

a projectile is launched with an initial speed of 58.5 m/s at an angle of 29.7 above the horizontal. the projectile lands on a hillside 3.8 s later. neglect air - friction. (assume that the x - axis is to the right and the y - axis is up along the page.) what is the straight - line distance from where the projectile was launched to where it hits its target? express your answer in meters.

Explanation:

Step1: Find the horizontal - component of the initial velocity

The initial velocity is $v_0 = 58.5$ m/s and the launch angle is $\theta$. The horizontal - component of the initial velocity is given by $v_{0x}=v_0\cos\theta$.

Step2: Use the horizontal - motion formula

The horizontal - motion formula is $x = v_{0x}t$. We know the time of flight $t = 3.8$ s. First, find $v_{0x}=58.5\cos\theta$. Then, $x=(58.5\cos\theta)\times3.8$.
Assuming the launch angle $\theta$ is given (not in the problem - statement, but if we assume a general case), if we assume the launch is at an angle $\theta$ with initial speed $v_0 = 58.5$ m/s and time $t = 3.8$ s, the horizontal distance $x$ (the straight - line distance from the launch point) is given by $x = v_0\cos\theta\times t$.
If we assume the launch angle $\theta$ is known, we substitute the values. For example, if the launch is at an angle $\theta = 30^{\circ}$:
$v_{0x}=58.5\cos30^{\circ}=58.5\times\frac{\sqrt{3}}{2}\approx50.6$ m/s.
$x = v_{0x}t=50.6\times3.8 = 192.28$ m.

Answer:

The horizontal distance $x = v_0\cos\theta\times t$, where $v_0 = 58.5$ m/s, $t = 3.8$ s and $\theta$ is the launch angle. If $\theta = 30^{\circ}$, $x\approx192.28$ m.