Question

a projectile is launched at a 30° angle relative to the ground. the projectile has an initial velocity of 15 m/s and travels through the air for 2 seconds. the horizontal displacement of the projectile, rounded to the nearest hundredth, is m.

Explanation:

Step1: Find horizontal - component of initial velocity

The formula for the horizontal - component of the initial velocity $v_{0x}$ is $v_{0x}=v_0\cos\theta$, where $v_0 = 15$ m/s and $\theta = 30^{\circ}$. So $v_{0x}=15\cos30^{\circ}=15\times\frac{\sqrt{3}}{2}\approx15\times0.866 = 12.99$ m/s.

Step2: Calculate horizontal displacement

The formula for horizontal displacement $x$ is $x = v_{0x}t$, where $t = 2$ s. Substitute $v_{0x}\approx12.99$ m/s and $t = 2$ s into the formula, we get $x=12.99\times2 = 25.98$ m.

Answer:

$25.98$