Question

How can a body achieve its maximum range in a projectile motion?​

Answer 1

Explanation:

First, just to be clear, what is projectile motion? The typical definition is the motion of an object due only to the gravitational force (no air resistance, rockets or stuff). If you want a detailed discussion about projectile motion, check out this post. Otherwise, remember the key to projectile motion:

Projectile motion is like two 1-d kinematics problems that only have the time in common. The acceleration in the vertical direction is -g and the horizontal acceleration is zero.

Projectile Motion - no air resistance

Using the main ideas above and the kinematic equations (for constant acceleration), the following should be true:

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Note that I am assuming at t = 0 seconds, the initial positions are x0y0 along with the initial velocities. Also, I am using the typical convention that g = 9.8 N/kg = 9.8 m/s2 so that the acceleration in the y-direction is -g. But how far will an object go if it both starts and ends at the same y? Here is a diagram showing the launch velocity of some object.

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The goal here is to find the range (x - x0). To do that, I will first determine the time of motion using the y-direction. Remember, I know that the object starts and finishes at the same y. This gives:

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Quick check. What value of θ would give the largest time? Well, that would be when sin(θ) is the greatest - at a value of π/2 (90 degrees - you know, straight up). What about the units? (m/s) over (m/s2) does give units of time. Great. Now putting this expression in for the x-motion.

[LaTeXiT-1-2]

Unit check. (m2/s2) over (m/s2) does indeed give units of meters. Another check. What if I shoot the ball straight up (θ = π/2)? Well, cos(π/2) = 0, so this gives a horizontal range of 0 meters. Makes sense.

But the real question is: what angle for the maximum distance (for a given initial velocity). Clearly this range depends on the product of sine and cosine. Let me first pull out a trig identity. The product of sine and cosine (in general) is:

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I know what you are thinking: we actually are going to use a trig-identity? I thought we just had to derive these in high school as a punishment for all those spit wads we threw. Oh no. They are actually useful. For this trig-identity, θ = φ so that:

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The greatest the value of sin of anything can be is 1. What angle would this be at?

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Done. 45 degrees. Just like the textbook says. Oh, you don't like this? You are a visual learner? I can deal with that. Here is a plot you might like. This is a plot of sinθ cosθ and the product of the two from zero to π/2.

Answer 2

Answer: $\alpha$=45°

Explanation:

• In a projectile motion the body covers distance on both x and y axes.
• The distance covered in x direction is known as range of the projectile motion.
• The formulae to calculate the range is= $\frac{v^{2} }{g}Sin \ 2\alpha$.

Where v= velocity at which the projectile is shot,

             g= acceleration due to gravity and

             $\alpha$= angle between the ground and the velocity vector.

• It will attain maximum velocity  when the value of sin will be maximum at $\alpha$=45°.
• Hence, it will be maximum at $\alpha$=45°.

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