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Exercises

Which of the following is/ are independent of the angle of projection of a projectile :

(a) time of flight

(b) maximum height reached

(c) acceleration of projectile

(d) horizontal component of velocity

The time of flight is determined by considering vertical motion. It means that time of flight is dependent on speed and the angle of projection.

T = 2 u y g = 2 u sin θ g

Maximum height is also determined, considering vertical motion. As such, maximum height also depends on the angle of projection.

H = u y 2 2 g = u 2 sin 2 θ 2 g

Horizontal component, being component of velocity, depends on the angle of projection.

u y = u cos θ

It is only the acceleration of projectile, which is equal to acceleration due to gravity and is, therefore, independent of the angle of projection. Hence, option (c) is correct.

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Two particles are projected with same initial speeds at 30° and 60° with the horizontal. Then

(a) their maximum heights will be equal

(b) their ranges will be equal

(c) their time of flights will be equal

(d) their ranges will be different

The maximum heights, ranges and time of lights are compared, using respective formula as :

(i)Maximum Height

H 1 H 2 = u 2 sin 2 θ 1 u 2 sin 2 θ 2 = sin 2 30 0 sin 2 60 0 = 1 2 2 3 2 2 = 1 3

Thus, the maximum heights attained by two projectiles are unequal.

(ii) Range :

R 1 R 2 = u 2 sin 2 θ 1 u 2 sin 2 θ 2 = sin 2 60 0 sin 2 120 0 = 3 2 2 3 2 2 = 1 1

Thus, the ranges of two projectiles are equal.

(iii) Time of flight

T 1 T 2 = 2 u sin θ 1 2 u sin θ 2 = sin 30 0 sin 60 0 = 1 2 3 2 = 1 3

Thus, the times of flight of two projectiles are unequal.

Hence, option (b) is correct.

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The velocity of a projectile during its flight at an elevation of 8 m from the ground is 3 i - 5 j in the coordinate system, where x and y directions represent horizontal and vertical directions respectively. The maximum height attained (H) by the particle is :

a 10.4 m b 8.8 m c 9.25 m d 9 m

We note that vertical component is negative, meaning that projectile is moving towards the ground. The vertical component of velocity 8 m above the ground is

v y = - 5 m / s

The vertical displacement (y) from the maximum height to the point 8 m above the ground as shown in the figure can be obtained, using equation of motion.

Projectile motion

Projectile motion

v y 2 = u y 2 + 2 a y

Considering the point under consideration as origin and upward direction as positive direction.

- 5 2 = 0 + 2 X - 10 X h h = - 25 / 20 = - 1.25 m

Thus, the maximum height, H, attained by the projectile is :

H = 8 + 1.25 = 9.25 m

Hence, option (c) is correct.

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A projectile is thrown with a given speed so as to cover maximum range (R). If "H" be the maximum height attained during the throw, then the range "R" is equal to :

a 4 H b 3 H c 2 H d H

The projectile covers maximum range when angle of projection is equal to 45°. The maximum range "R" is given by :

R = u 2 sin 2 θ g = u 2 sin 90 0 g = u 2 g

On the other hand, the maximum height attained by the projectile for angle of projection, 45°, is :

H = u y 2 2 g = u 2 sin 2 45 0 2 g = u 2 4 g

Comparing expressions of range and maximum height, we have :

R = 4 H

Hence, option (a) is correct.

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The speed of a projectile at maximum height is half its speed of projection, "u". The horizontal range of the projectile is :

a 3 u 2 g b 3 u 2 2 g c u 2 4 g d u 2 2 g

The horizontal range of the projectile is given as :

R = u 2 sin 2 θ g

In order to evaluate this expression, we need to know the angle of projection. Now, the initial part of the question says that the speed of a projectile at maximum height is half its speed of projection, "u". However, we know that speed of the projectile at the maximum height is equal to the horizontal component of projection velocity,

u cos θ = u 2 cos θ = 1 2 = cos 30 0 θ = 30 0

The required range is :

R = u 2 sin 2 θ g = u 2 sin 60 0 g = 3 u 2 2 g

Hence, option (b) is correct.

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Let " T 1 ” and “ T 2 ” be the times of flights of a projectile for projections at two complimentary angles for which horizontal range is "R". The product of times of flight, " T 1 T 2 ”, is equal to :

a R g b R 2 g c 2 R g d R 2 g

Here, we are required to find the product of times of flight. Let " θ " and " 90 0 θ " be two angles of projections. The times of flight are given as :

T 1 = 2 u sin θ g

and

T 2 = 2 u sin 90 0 θ g = 2 u cos θ g

Hence,

T 1 T 2 = 4 u 2 sin θ cos θ g 2

But,

R = 2 u 2 sin θ cos θ g

Combining two equations,

T 1 T 2 = 2 R g

Hence, option (c) is correct.

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A projectile is projected with a speed "u" at an angle " θ " from the horizontal. The magnitude of average velocity between projection and the time, when projectile reaches the maximum height.

a u cos 2 θ b 3 u 2 cos θ + 1

c 3 u 2 cos θ + 1 2 u d u 3 u 2 cos θ + 1 2

The magnitude of average velocity is given as :

v a v g = Displacement Time

Now the displacement here is OA as shown in the figure. From right angle triangle OAC,

Projectile motion

Projectile motion

O A = O C 2 + B C 2 = { R 2 2 + H 2 } O A = R 2 4 + H 2

Hence, magnitude of average velocity is :

v a v g = R 2 4 + H 2 T

Substituting expression for each of the terms, we have :

v a v g = u 4 sin 2 2 θ 4 g 2 + u 4 sin 4 θ 4 g 2 2 u sin θ g

vavg = u(3 u2cos + 1)/2 v a v g = u 3 u 2 cos θ + 1 2

Hence, option (d) is correct.

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A projectile is projected at an angle " θ " from the horizon. The tangent of angle of elevation of the highest point as seen from the position of projection is :

a tan θ 4 b tan θ 2 c tan θ d 3 tan θ 2

The angle of elevation of the highest point " θ " is shown in the figure. Clearly,

Projectile motion

Projectile motion

tan α = A C O C = H R 2 = 2 H R

Putting expressions of the maximum height and range of the flight, we have :

tan α = 2 u 2 sin 2 θ X g 2 g X u 2 sin 2 θ = sin 2 θ 2 sin θ cos θ = tan θ 2

Hence, option (b) is correct.

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In a firing range, shots are taken at different angles and in different directions. If the speed of the bullets is "u", then find the area in which bullets can spread.

a 2 π u 4 g 2 b u 4 g 2 c π u 4 g 2 d π u 2 g 2

The bullets can spread around in a circular area of radius equal to maximum horizontal range. The maximum horizontal range is given for angle of projection of 45°.

R max = u 2 sin 2 θ g = u 2 sin 2 X 45 0 g = u 2 sin 90 0 g = u 2 g

The circular area corresponding to the radius equal to maximum horizontal range is given as :

A = π R max 2 = π X u 2 g 2 = π u 4 g 2

Hence, option (c) is correct.

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More exercises

Check the module titled " Features of projectile motion (application) to work out more problems.

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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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