Question

The trajectory of a bullet is given by $x=v_0(\cos \alpha )$ ty $=v_0(\sin \alpha )t-\frac{1}{2}g{t}^2$ where $v_0=500\mathrm{m}/\mathrm{s}$, $g=9.8=9.8\mathrm{m}/{\mathrm{s}}^2$, and $\alpha =30$ degrees. When will the bullet hit the ground? How far from the gun will the bullet hit the

Short answer

The bullet hits the ground after approximately 51.02 seconds and lands approximately 22104.7 meters from the gun.

Step-by-step solution

Understand the Equation

The trajectory equations are given for horizontal position $x$ and vertical position $y$ in terms of time $t$. We have: $$x=v_0\cos (\alpha )\cdot t$$ and $$y=v_0\sin (\alpha )\cdot t-\frac{1}{2}g{t}^2.$$ For the bullet to hit the ground, the vertical position $y$ must be zero.

Set Vertical Position to Zero

We need to find when $y=0$ in the equation: $$y=v_0\sin (\alpha )t-\frac{1}{2}g{t}^2.$$ Plug in the values of $v_0=500$, $g=9.8$, and $\alpha ={30}^{\circ }$. This gives: $$0=500\sin ({30}^{\circ })t-\frac{1}{2}\cdot 9.8\cdot t^2.$$

Simplify Using Trigonometric Values

The sine of 30 degrees is 0.5. Substitute this into the equation to simplify it: $$0=500\cdot 0.5\cdot t-4.9t^2$$ which simplifies to: $$0=250t-4.9t^2.$$

Solve for Time t

Factor out $t$ from the equation:$$0=t(250-4.9t).$$ This implies $t=0$ or $250-4.9t=0$. Since $t=0$ is the initial time, solve for $t$: $$4.9t=250$$$$t=\frac{250}{4.9}\approx 51.02$$ seconds.

Calculate Horizontal Distance

Now that we know $t=51.02$ seconds, substitute into the horizontal position equation:$$x=v_0\cos (\alpha )\cdot t.$$ Using $\cos ({30}^{\circ })=\sqrt{3}/2\approx 0.866$:$$x=500\cdot 0.866\cdot 51.02.$$ Calculate $x$: $$x\approx 22104.7$$ meters.

Key concepts

Trajectory Calculations

Calculating the trajectory of a projectile is crucial in understanding its path. The trajectory can be described by two equations, each representing movement in either the horizontal or the vertical direction. The equation for horizontal motion is given by:

• $$x=v_0\cos (\alpha )t$$

This equation explains how the horizontal distance covered by the bullet varies with time $t$. Here, $v_0$ is the initial velocity of the bullet, and $\alpha$ is the angle of projection. The horizontal coordinate $x$ is directly proportional to time when considering consistent speeds.
The vertical motion follows a slightly more complex path due to gravity:

• $$y=v_0\sin (\alpha )t-\frac{1}{2}g{t}^2$$

This equation includes gravitational acceleration $g$ acting on the bullet, which influences how far the bullet rises and falls across time. This vertical component equation is essential for determining when and where the bullet will return to the ground.

Trigonometry in Physics

Trigonometry plays a vital role in resolving projectile motion equations. It helps break down the initial velocity $v_0$ into horizontal and vertical components using the angle of projection $\alpha$. For a projectile like a bullet fired at a 30-degree angle, these trigonometric functions become pivotal:

• The cosine function $\cos (\alpha )$ calculates the horizontal component: $$v_0\cos ({30}^{\circ })\approx v_0\times 0.866$$
• The sine function $\sin (\alpha )$ finds the vertical component:$$v_0\sin ({30}^{\circ })=v_0\times 0.5$$

Using these components, one can determine how far and how high the projectile will travel over time. Trigonometry, hence, bridges the gap between angle, velocity, and motion, giving us a comprehensive picture of the projectile's trajectory.

Horizontal and Vertical Motion

In projectile motion, understanding horizontal and vertical motion separately can clarify a lot about the overall movement.
The horizontal motion of the projectile is uniform, meaning it travels in a straight line at a constant speed since there is no acceleration affecting it (ignoring air resistance). This is captured by the equation:

• $$x=v_0\cos (\alpha )t$$

Conversely, vertical motion, affected by gravity, is non-uniform, as portrayed in:

• $$y=v_0\sin (\alpha )t-\frac{1}{2}g{t}^2$$

Here, gravity introduces a downward acceleration of 9.8 m/s² that reduces the vertical velocity until it becomes zero at the peak of the trajectory, and then pulls the bullet back down. Hence, the vertical and horizontal motions together define the projectile's entire path.

Kinematics Equations

Kinematics equations in projectile motion articulate how various factors affect an object's position and velocity over time. In this exercise, two principal equations are involved.
The vertical motion equation:

• $$y=v_0\sin (\alpha )t-\frac{1}{2}g{t}^2$$

This equation outlines how vertical position changes with time, factoring in the initial vertical velocity and gravitational force.
The horizontal motion, on the other hand, is expressed by:

• $$x=v_0\cos (\alpha )t$$

This states that horizontal distance depends on constant horizontal velocity, unaffected by acceleration in ideal conditions. Together, these kinematics equations allow you to predict when and where the projectile will land, with the ability to calculate the exact time it hits the ground and the range it will cover.