Question

Use the model for projectile motion, assuming there is no air resistance. \([a(t)=-9.8\) meters per second per second \(]\) A projectile is fired from ground level at an angle of \(8^{\circ}\) with the horizontal. The projectile is to have a range of 50 meters. Find the minimum velocity necessary.

Short answer

To solve this problem, use the formula for the range of a projectile, rearrange it to solve for the initial speed, and then plug in the given values. This yields the minimum initial speed needed for the projectile to achieve a range of 50 m when launched at an angle of 8 degrees.

Step-by-step solution

Understand the problem and select the correct equation

The range \(R\) of a projectile launched with an initial speed \(v\) at an angle \(\(\theta\)\) (assuming no air resistance) is given by the formula \[R = \frac{{v^2}}{g} sin(2\theta)\], where \(g\) is the acceleration due to gravity. In this problem, we are given \(R = 50 m\), \(\theta = 8^o\), and \(g = -9.8 m/s^2\), and we want to solve for the minimum initial speed \(v\). Thus, we need to rearrange the above equation to solve for \(v\).

Rearrange the formula and solve for v

By rearranging the formula for range, we find \[v = \sqrt{{Rg \over sin(2\theta)}}\]. From here, plug in the given values and compute the result. Make sure to convert the angle from degrees to radians before making the calculation, as the trigonometric functions typically expect the argument to be in radians.

Compute the Minimum Initial Speed

After plugging the given values into the equation, the next step is to calculate the quantity square-root of \(\frac{{(50 m) (-9.8 m/s^2)}}{sin(2 * (8 degrees))}\) to obtain the minimum initial speed. Make sure to take the positive root, as speed cannot be negative.

Key concepts

Initial Velocity Calculation

When we discuss the initial velocity of a projectile, we are referring to the speed at which the object is launched into the air. Determining the minimum initial velocity is crucial to ensure the projectile achieves a certain goal, such as reaching a specific range.

To begin the calculation, we consider the formula for the range of a projectile, which states that the range (\(R\text{ m}\)) is proportional to the square of the initial velocity (\(v\text{ m/s}\)) and the sine of twice the launch angle (\(2\theta\text{ degrees}\)), all over the acceleration due to gravity (\(g\text{ m/s}^2\text{, usually }-9.8\text{ m/s}^2\text{ on Earth}\)). The formula is represented as \[R = \frac{{v^2}}{g} \sin(2\theta)\] Understanding this equation allows us to derive the initial velocity, which involves rearranging the equation to isolate and solve for \(v\).

To solve for \(v\), it's essential to invert the formula to \[v = \sqrt{{Rg \over \sin(2\theta)}}\] Keep in mind, angles must be in radians when inserting them into trigonometric functions, which requires a conversion from degrees if necessary. If we place our known values into the rearranged formula, we calculate the minimum initial velocity specific for our projectile's range.

Range of a Projectile

The range of a projectile is the horizontal distance it covers before it returns to the same elevation from which it was launched—in this case, ground level. The range is an important factor to consider in projectile motion as it helps determine how far the projectile will travel.

The equation \[R = \frac{{v^2}}{g} \sin(2\theta)\] not only assists in calculating initial velocity but also can be used directly to predict the range when the initial velocity and angle of launch are known. When we fix the angle and gravity, the range is maximized at a 45-degree launch angle; any greater or lesser will shorten the distance traveled.

While the concept might appear straightforward, executing the calculation requires careful consideration of units and conversions. For example, angles must be converted from degrees to radians to utilize trigonometric functions correctly. Moreover, it requires a clear understanding of how gravity, velocity, and angle work together to shape a projectile’s trajectory.

Acceleration Due to Gravity

Acceleration due to gravity is a constant force that acts on objects upon release, pulling them towards the earth's center. On the surface of the Earth, this acceleration is approximately \(9.8\text{ m/s}^2\), and in physics problems, it is often represented as a negative value to indicate that it acts in the opposite direction of the rising projectile.

This force is crucial in the equations of projectile motion and determines how quickly an object will fall after reaching the peak of its trajectory. The acceleration due to gravity affects every aspect of the projectile's path from the moment it is launched to when it lands.

It's also essential to remember that gravity acts only on the vertical motion of the projectile and does not influence the horizontal motion. This separation of vertical and horizontal components allows us to analyze projectile motion conveniently within the gravity's context and calculate the trajectory and final position of the object in motion.