# 5.2 Features of projectile motion  (Page 6/7)

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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.

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,

$\begin{array}{l}y=x\mathrm{tan}\theta -\frac{g{x}^{2}}{2{u}^{2}{\mathrm{cos}}^{2}\theta }\end{array}$

$\begin{array}{l}⇒15=30\mathrm{tan}{45}^{0}-\frac{10x{30}^{2}}{2{u}^{2}{\mathrm{cos}}^{2}{45}^{0}}\\ ⇒15=30x1-\frac{10x2x{30}^{2}}{2{u}^{2}}\\ ⇒15=30-\frac{9000}{{u}^{2}}\\ ⇒{u}^{2}=600\\ ⇒u=24.49\phantom{\rule{2pt}{0ex}}m/s\end{array}$

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 :

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

$\begin{array}{l}y=x\mathrm{tan}\theta -\frac{g{x}^{2}\left(1+{\mathrm{tan}}^{2}\theta \right)}{2{u}^{2}}\end{array}$

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 :

$⇒y=x\mathrm{tan}\theta -\frac{g{x}^{2}}{2{u}^{2}}\left(1+{\mathrm{tan}}^{2}\theta \right)$

Putting values,

$⇒13=50\mathrm{tan}\theta -\frac{10x{50}^{2}}{2x{\left(10\sqrt{g}\right)}^{2}}\left(1+\mathrm{tan}{}^{2}\theta \right)$

$⇒25{\mathrm{tan}}^{2}\theta -100\mathrm{tan}\theta +51=0$

$⇒\mathrm{tan}\theta =17/5\phantom{\rule{1em}{0ex}}or\phantom{\rule{1em}{0ex}}3/5$

Taking the smaller angle of projection to hit the target,

$⇒\theta ={\mathrm{tan}}^{-1}\left(\frac{3}{5}\right)$

## 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 :

$\begin{array}{l}\frac{{u}^{2}{\mathrm{sin}}^{2}\theta }{\mathrm{2g}}=\frac{{u}^{2}\mathrm{sin}2\theta }{g}\end{array}$

$\begin{array}{l}{u}^{2}{\mathrm{sin}}^{2}\theta =2{u}^{2}\mathrm{sin}2\theta \\ ⇒{\mathrm{sin}}^{2}\theta =2\mathrm{sin}2\theta =4\mathrm{sin}\theta x\mathrm{cos}\theta \\ ⇒\mathrm{sin}\theta =4\mathrm{cos}\theta \\ ⇒\mathrm{tan}\theta =4\\ ⇒\theta ={\mathrm{tan}}^{-1}\left(4\right)\end{array}$

#### Questions & Answers

How can we take advantage of our knowledge about motion?
Kenneth Reply
pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
Mercy Reply
what are scalars
Abdhool Reply
show that 1w= 10^7ergs^-1
Lawrence Reply
what's lamin's theorems and it's mathematics representative
Yusuf Reply
if the wavelength is double,what is the frequency of the wave
Ekanem Reply
What are the system of units
Jonah Reply
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
Samson Reply
58asagravitasnal firce
Amar
water boil at 100 and why
isaac Reply
what is upper limit of speed
Riya Reply
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
Clash Reply
which colour has the shortest wavelength in the white light spectrum
Mustapha Reply
how do we add
Jennifer Reply
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
Abhyanshu Reply
x=5.8-3.22 x=2.58
sajjad
what is the definition of resolution of forces
Atinuke Reply

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