Question

multiple choice 2 points a dart is shot at an angle of 35 degrees to the horizontal. if the dart is in the air for 3 seconds and lands 45 m away, what was the initial velocity of the dart? 15.1 m/s 30.1 m/s 25.2 m/s 18.3 m/s multiple choice 2 points a cannon is shot, from ground level, at an angle of 30 degrees to the horizontal. if the initial velocity of the cannon - ball is 200 m/s, and it also lands at the same height (ground level) from which it was shot, how far did the cannon - ball land? 4464 meters 1464 meters 2464 meters 3464 meters

Explanation:

Step1: Analyze horizontal - motion formula for the dart

The horizontal - motion formula is $x = v_{0x}t$, where $x$ is the horizontal displacement, $v_{0x}$ is the initial horizontal velocity, and $t$ is the time of flight. The initial horizontal velocity $v_{0x}=v_0\cos\theta$, so $x = v_0\cos\theta\times t$. We know $x = 45$ m, $\theta = 35^{\circ}$, and $t = 3$ s.

Step2: Solve for the initial velocity $v_0$ of the dart

Rearranging the formula $x = v_0\cos\theta\times t$ for $v_0$, we get $v_0=\frac{x}{t\cos\theta}$. Substitute $x = 45$ m, $t = 3$ s, and $\theta = 35^{\circ}$ ($\cos35^{\circ}\approx0.819$) into the formula: $v_0=\frac{45}{3\times0.819}=\frac{45}{2.457}\approx18.3$ m/s.

Step3: Analyze horizontal - motion formula for the cannon - ball

For the cannon - ball, the range formula when the initial and final heights are the same is $R=\frac{v_0^2\sin2\theta}{g}$, where $v_0 = 200$ m/s, $\theta = 30^{\circ}$, and $g = 9.8$ m/s². First, find $\sin2\theta=\sin(2\times30^{\circ})=\sin60^{\circ}=\frac{\sqrt{3}}{2}\approx0.866$.

Step4: Calculate the range of the cannon - ball

Substitute the values into the range formula $R=\frac{v_0^2\sin2\theta}{g}$. $R=\frac{200^{2}\times0.866}{9.8}=\frac{40000\times0.866}{9.8}=\frac{34640}{9.8}\approx3534.69$ m, which is closest to 3464 meters.

Answer:

1. A. 18.3 m/s
2. A. 3464 meters