Projectile Motion

Oblique Projection
1
1. x= u cosθ t & y= u sinθ t − gt 2
2
2. Equation of trajectory:
g
y = x tanθ − ( 2 2 )x 2
2u cos θ
3. Velocity v after certain time
𝑣𝑥 =u cosθ &𝑣𝑦 = u sinθ -gt
Resultant velocity v=√𝑣𝑥 2 + 𝑣𝑦 2
𝑣𝑦 u sinθ−gt
=√u2 − 2ugt sinθ + g 2 t 2 at angle 𝛼 = 𝑡𝑎𝑛−1 ( ) = 𝑡𝑎𝑛−1 ( )
𝑣 𝑥 u cos θ
4. Equation of trajectory:
x
y = x tanθ (1 - )
R
5. Time to attain greatest vertical height:
u sinθ uy R 2H T
t= = = =√ =
g g 2u cosθ g 2
6. Time of flight:
2u sinθ 2uy R 2H
T= 2t = = = = 2√
g g u cosθ g
7. Horizontal Range:
u2 sin 2θ 2ux uy 4u2 sin2 θ g cosθ g
R=ux T = = = 4H cotθ = × = T 2 cotθ
g g g2 2 sinθ 2
8. Maximum height:
u2 sin2 θ uy 2 1 1 1
H= = = gt 2 = gT 2 = R tanθ
2g 2g 2 8 4
 For maximum horizontal range, angle of projection should be 45°, and
u2
maximum range= .
g
 R= H when θ = tan-1(4) =75.96°.
Horizontal Projection
1
1. x= ut & y= gt 2
2
2. Equation of trajectory:
g
y = ( 2)x 2
2u
 3. Velocity v after certain time
𝑣𝑥 =u &𝑣𝑦 =gt
Resultant velocity v=√𝑣𝑥 2 + 𝑣𝑦 2
𝑣𝑦 gt
=√u2 + g 2 t 2 at angle 𝛼 = 𝑡𝑎𝑛−1 ( ) = 𝑡𝑎𝑛−1 ( )
𝑣 𝑥 u
4. Time of flight:
2ℎ
T=√
𝑔
5. Horizontal Range:
2ℎ
R=ux T=u√
𝑔

6. Final velocity vf=√u2 + 2gh

2gh gT
=√u2 + g 2 T 2 at angle 𝛼 = 𝑡𝑎𝑛−1 ( ) = 𝑡𝑎𝑛−1 ( u )
u

Questions:
1. An object is projected from a point on the ground level with a velocity
60𝑖̂+20√3𝑗̂. Calculate its time of flight, maximum height and horizontal
range.
2. A projectile is thrown with an initial velocity 20√3𝑖̂+20𝑗̂ on the horizontal
plane. Calculate its time of flight, maximum height and horizontal range.
3. A baseball is thrown towards a player with an initial speed of 𝟐𝟎 𝒎𝒔−𝟏 and
𝟒𝟓𝒐 with the horizontal. At the moment the ball is thrown, the player is 50 m from
the thrower. At what speed and in what direction must he run to catch the ball at the
same height at which it was released? (g = 10 𝒎𝒔−𝟐 )
Ans.: 3.5 𝒎𝒔−𝟏
4. A body is projected with a velocity of 40 𝒎𝒔−𝟏 . After two seconds it crosses a
vertical pole of height 20.4m. Calculate the angle of projection and horizontal
range of projectile. (g = 10 𝒎𝒔−𝟐 )
Ans.: 𝟑𝟎𝒐 , 141.4m
5. A batter hits a baseball so that it leaves the bat with an initial speed of
37m/s at the angle of 𝟓𝟑. 𝟏𝒐 , at a location where g = 10 𝒎𝒔−𝟐 . (a) Find the
position of the ball and the magnitude and direction of its velocity after 2
seconds. (Ans.-39.6 m, 𝟐𝟒. 𝟐𝒐 ) (b) Find the time when the ball reaches the
highest point of its flight and find its height h at this point. (Ans.- 3.02 s,
44.7 m)(c) Find the horizontal range (Ans. 134 m).
6. A body is projected with a speed of 30ms−1 at an angle of 𝟑𝟎𝒐 with the vertical.
Find the maximum height, time of flight and the horizontal range of the motion.
[Take g = 10 𝒎𝒔−𝟐 ]
Ans.: (Height-34.44 m, time- 5.3 s, range 79.53 m)
7. A particle at a height 'h' from the ground is projected with an angle 30∘ from the
horizontal, it strikes the ground making angle 45∘ with horizontal. It is again
projected from the same point at height h with the same speed but with an angle
of 60∘ with horizontal. Find the angle it makes with the horizontal when it strikes
the ground. [Take g= 10 𝒎𝒔−𝟐 ]
Ans.: 𝒕𝒂𝒏−𝟏 √𝟓
8. A projectile has a range of 50m and reaches a maximum height of 10m calculate
the angle at which the projectile is fired.
Ans.: 𝟑𝟖. 𝟔𝟔𝒐
9. A boy stands at 39.2 m from a building and throws a ball which just passes
through a window 19.6 m above the ground. Calculate the velocity of
projection of the ball.
Ans.: 27.72 𝒎𝒔−𝟏
10.Q. 8. Find the angle of projection at which horizontal range and maximum
height are equal.
Ans.: 𝟕𝟓𝒐 𝟓𝟖′
11. Q. 9. Prove that the maximum horizontal range of four times the maximum height
attained by a projectile which is fired along the required oblique direction.
12. A ball is kicked at an angle of 30° with the vertical. If
the horizontal component of its velocity is 19.6 m/s, what is its maximum
height and the horizontal range?
Ans.: 58.5 m & 135.8 m
13.Prove that a gun will shoot three times as high when its angle of elevation 60
degree as when it is 30 degree, but cover the same horizontal range.
14. A ball is thrown at angle θ and another ball is thrown at an angle (90−θ) with the
horizontal direction from the same point with the same speed 40𝒎𝒔−𝟏 . The second
ball reaches 50m higher than the first ball. Find their individual heights.
Ans. 15 m and 65 m
15.If R is the horizontal for θ inclination and h is the maximum height reached
𝑹𝟐
by a projectile, show that its maximum range is given by ( + 𝟐𝒉).
𝟖𝒉
16. There are two angles of projection for which the horizontal range is the same.
Show that the sum of the maximum heights for these two angles is independent of
the angles of projection.
17. Show that there are two values of time for which a projectile is at the same height.
Also show mathematically that the sum of these two times is equal to the time of
flight.
18. Two projectiles are thrown with different velocities and at different angles so as to
cover the same maximum height. Show that the sum of the times taken by each to
reach the highest point is equal to the total time taken by either of the projectiles.
19.A hunter aims his gun and fires a bullet directly at a monkey on a tree. At
the instant the bullet leaves the barrel of the gun, the monkey drops. Will the
bullet hit the monkey? Explain your answer.
(Ans.: The bullet and the monkey will reach at a point of same height at
the same time. Hence, the bullet will always the monkey whatever be the
velocity of the bullet.)
(1) A ball is projected with an initial upward velocity component of 20 𝑚𝑠 −1 and a horizontal
velocity component 25 𝑚𝑠 −1 (g= 9.8 𝑚𝑠 −2).
(a) Find the position and velocity of the ball after 4 sec.
(b) How much time is required to reach the highest point of the trajectory?
(c) How high is this point?
(d) How much time (after launch) is required for the ball to return to its original level?
(e) How far has it travelled horizontally during this time?

Illustrate your answer with neat and sufficiently large sketch.

[Ans.: (a) x=100 m, y=1.6 m, 𝒗𝒙 = 𝟐𝟓 𝒎𝒔−𝟏 , 𝒗𝒚 = −𝟏𝟗. 𝟐 𝒎𝒔−𝟏 (b) 2.04 s (c) 20.4 m
(d) 4.08 s (e) 102 m].
(2) A bullet is projected with a velocity 196 𝑚𝑠 −1 at an angle 30𝑜 with the horizontal; find (i)
the greatest height attained (ii) the range on the horizontal plane (iii) the time of flight (iv)
the velocity and the direction of the motion of the bullet when it is at height of 294 m
Ans.: (i) 490 m (ii) 3394.8 m (iii) 20 sec (iv) 180.7𝒎𝒔−𝟏 , at an angle of 𝒕𝒂𝒏−𝟏 𝟎. 𝟑𝟔]
(3) Find the velocity and direction of projection of a shot which passes in a horizontal direction
just over the top of a wall which is 2 m off and 19.6 m high.
Ans.: 27.7 𝑚𝑠 −1 at an angle 45𝑜 with the horizontal].
(4) The time of flight in a horizontal plane through the point of projection is 6 seconds; find the
greatest height attained. [ Ans.: 44.1 m ??]
(5) Prove that projectile will rise three times as high when its angle of elevation is 60𝑜 as when
it is 30𝑜 , but will cover the same horizontal distance.
(6) Find the angle of projection for which the maximum vertical height and the horizontal wanre
are same. [Ans.: 𝟕𝟔𝒐 ]
(7) A projectile is fired with a horizontal velocity of 30 𝑚𝑠 −1 from the top of a cliff 80 m high.
(a) How long will it take to strike the level ground at the base of the cliff? (b) How far from
the foot of the cliff will it strike? (c) With what velocity will it strike? (g=10 𝑚𝑠 −2 ).
𝟒
[ 𝑨𝒏𝒔. : (𝒂)𝟒 𝐬𝐞𝐜(𝒃) 𝟏𝟐𝟎 𝒎 (𝒄)𝟓𝟎 𝒎𝒔−𝟏 𝒂𝒕 𝒕𝒂𝒏−𝟏 (𝟑)].
(8) An airplane flying horizontally at 320 𝑚𝑠 −1 releases a bomb at a height of 1600 m. Find
the time take for the bomb to reach the ground and also magnitude and direction of the
velocity with which it strikes the ground. [Ans.: 10 sec, 320.2 𝒎𝒔−𝟏 at 𝟒𝟓𝒐 to ground]
(9) With what velocity must a stone be projected horizontally from the top of a tower 78.4 m
high so as to reach a point on the ground 400 meters from the foot of the tower?
(𝑔 = 9.8 𝑚𝑠 −2 ). [Ans.:100 𝒎𝒔−𝟏 ]
(10) A helicopter flying horizontally with a speed of 30 𝑚𝑠 −1 at an altitude of 500 m, has to drop
food packet for a person standing on the ground. At what distance from the person should
the packet be dropped? The man stands in the vertical plane of the helicopter’s motion.
(𝑔 = 10 𝑚𝑠 −2). [Ans.:300 m]
(11) A player kicks a football at an angle of 37𝑜 with the horizontal and with an initial speed
of 316 𝑚𝑠 −1 . A second player at a distance of 33 m from the first in the direction of kick
starts running to meet the ball at the instant it is kicked. How fast must he run in order to
catch the ball before it hits the ground? [Ans.: 44 m]