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Hitting a specified target

An archer aims a bull’s eye; a person throws a pebble to strike an object placed at height and so on. The motion involved in these situations is a projectile motion – not a straight line motion. The motion of the projectile (arrow or pebble) has an arched trajectory due to gravity. We need to aim higher than line of sight to the object in order to negotiate the loss of height during flight.

Projectile motion

The projectile hits the wall at a given point.

Hitting a specified target refers to a target whose coordinates (x,y) are known. There are two different settings of the situation. In one case, the angle of projection is fixed. We employ equation of the projectile to determine the speed of projectile. In the second case, speed of the projectile is given and we need to find the angle(s) of projection. The example here illustrates the first case.

Problem : A projectile, thrown at an angle 45° from the horizontal, strikes a building 30 m away at a point 15 above the ground. Find the velocity of projection.

Solution : As explained, the equation of projectile path suits the description of motion best. Here,

x = 30 m, y = 15 m and θ = 45°. Now,

y = x tan θ - g x 2 2 u 2 cos 2 θ

15 = 30 tan 45 0 - 10 x 30 2 2 u 2 cos 2 45 0 15 = 30 x 1 - 10 x 2 x 30 2 2 u 2 15 = 30 - 9000 u 2 u 2 = 600 u = 24.49 m / s

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As pointed out earlier, we may need to determine the angle(s) for a given speed such that projectile hits a specified target having known coordinates. This presents two possible angles with which projectile can be thrown to hit the target. This aspect is clear from the figure shown here :

Projectile motion

Projectile motion

Clearly, we need to use appropriate form of equation of motion which yields two values of angle of projection. This form is :

y = x tan θ - g x 2 ( 1 + tan 2 θ ) 2 u 2

This equation, when simplified, form a quadratic equation in "tanθ". This in turn yields two values of angle of projection. Smaller of the angles gives the projection for least time of flight.

Problem : A person standing 50 m from a vertical pole wants to hit the target kept on top of the pole with a ball. If the height of the pole is 13 m and his projection speed is 10√g m/s, then what should be the angle of projection of the ball so that it strikes the target in minimum time?

Solution : Equation of projectile having square of “tan θ” is :

y = x tan θ g x 2 2 u 2 1 + tan 2 θ

Putting values,

13 = 50 tan θ 10 x 50 2 2 x 10 g 2 1 + tan 2 θ

25 tan 2 θ - 100 tan θ + 51 = 0

tan θ = 17 / 5 o r 3 / 5

Taking the smaller angle of projection to hit the target,

θ = tan - 1 3 5

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Determining attributes of projectile trajectory

A projectile trajectory under gravity is completely determined by the initial speed and the angle of projection or simply by the initial velocity (direction is implied). For the given velocity, maximum height and the range are unique – notably independent of the mass of the projectile.

Thus, a projectile motion involving attributes such as maximum height and range is better addressed in terms of the equations obtained for the specific attributes of the projectile motion.

Problem : Determine the angle of projection for which maximum height is equal to the range of the projectile.

Solution : We equate the expressions of maximum height and range (H = R) as :

u 2 sin 2 θ 2g = u 2 sin 2 θ g

u 2 sin 2 θ = 2 u 2 sin 2 θ sin 2 θ = 2 sin 2 θ = 4 sin θ x cos θ sin θ = 4 cos θ tan θ = 4 θ = tan - 1 ( 4 )

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Source:  OpenStax, Kinematics fundamentals. OpenStax CNX. Sep 28, 2008 Download for free at http://cnx.org/content/col10348/1.29
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