## A particle is projected from the ground level. It just passes through upper ends of vertical poles ….

Q: A particle is projected from the ground level. It just passes through upper ends of vertical poles A, B, C of height 20 m, 30 m and 20 m respectively. The time taken by the particle to travel from B to C is double of the time taken from A to B. Find the maximum height attained by the particle from the ground level.

Ans: 125/4 m

Solution: tAB = t
tBC = 2 t
So, for ABC part

tAC = 3 t $= \frac{2 u_y}{g}$

$u_y = \frac{3}{2}g t$

Also ; $10 = u_y – \frac{1}{2}g t^2 = g t^2$

t = 1 s

uy = 15

$h = \frac{u_y^2}{2 g} = \frac{225}{20} = \frac{45}{4}$

Maximum height attained $= 20 + \frac{45}{4} = \frac{125}{4} m$

## The plane of projectile motion passes through a horizontal line PQ which makes an angle of 37º with positive x-axis …

Q: The plane of projectile motion passes through a horizontal line PQ which makes an angle of 37º with positive x-axis, xy plane is horizontal. The coordinates of the point where the particle will strike the line PQ is: (Take g = 10 m/s2)

(A) (10, 6, 0)m

(B) (8, 6, 0)m

(C) (10, 8, 0)m

(D) (6, 10,0)m

Ans: (A)

Solution: Range = 10 m. For point where particle strikes line PQ

x coordinate = 10 cos 37° + 2 = 10 m

y coordinate = 10 sin 37° = 6 m

z coordinate = 0 m

## Several particles are thrown from a point in various directions but with the same initial speed vₒ…

Q: Several particles are thrown from a point in various directions but with the same initial speed vₒ. At any subsequent time t before they reach the ground, they lie on the surface of

(a) parabolic

(b) sphere

(c) ellipsoid

(d) straight line

Ans: (b)

## A stone is to be projected horizontally from top of a 1.7 m high pole. Calculate initial velocity of projection …

Q: A stone is to be projected horizontally from top of a 1.7 m high pole. Calculate initial velocity of projection (in ms-1), if strikes perpendicularly an inclined plans as shown in the figure.

Ans: 3

## A fountain consist of a small hemispherical sprayer which lies on the surface of water in a basin as shown in the figure…

Q: A fountain consist of a small hemispherical sprayer which lies on the surface of water in a basin as shown in the figure. The sprayer has in numerable small holes through which water spurts at same speed 10 ms-1 in all directions. Choose the correct option(s)

(a) The shape of water bell formed by the jets is parabolic

(b) The maximum height attained by the water is 5 m

(c) The diameter of the basin be at least 20 m If no water is to be lost

(d)  None of the above

Ans: (a) , (b) , (c)

Solution: Here Path of Projectile is parabolic , hence shape of water bell formed by the jets is parabolic

Maximum height ,$\displaystyle H = \frac{u^2}{2 g}$

$\displaystyle H = \frac{10^2}{2 \times 10}$

H = 5 m

Maximum Range , $\displaystyle R = \frac{u^2}{g}$

$\displaystyle R = \frac{10^2}{10}$

R = 10 m

Diameter D = 2R = 2 × 10 = 20 m

## Two stones A and B are thrown simultaneously from a point on horizontal ground. The stone A is thrown vertically up ….

Q: Two stones A and B are thrown simultaneously from a point on horizontal ground. The stone A is thrown vertically up with velocity uA and the stone B is thrown with speed uB at angle 37° with horizontal . After 0.9 sec stones are moving perpendicular to each other. Then,

(a) uA > 9ms-1

(b) uB > 9ms-1

(c) uB= 15ms-1

(d) uB= 9ms-1

Ans: (a) ,(c)

Sol: Time of Ascent for ball A $\displaystyle t_A = \frac{u_A}{g}$

$\displaystyle \frac{u_A}{g} > 0.9$

$\displaystyle \frac{u_A}{10} > 0.9$

$\displaystyle u_A > 9 m/s$

For ball B time of Ascent $\displaystyle t_B = \frac{u_B sin\theta}{g}$

$\displaystyle \frac{u_B sin\theta}{g} = 0.9$

$\displaystyle \frac{u_B sin37^o}{10} = 0.9$

$\displaystyle u_B sin37^o = 9$

$\displaystyle u_B \times \frac{3}{5} = 9$

$\displaystyle u_B = 15 m/s$

## A projectile thrown from ground with speed 15 ms-1 at an angle 37° with vertical. During motion,

Q: A projectile thrown from ground with speed 15 ms-1 at an angle 37° with vertical. During motion, (given, g=10 ms-2)

(a) minimum speed is 9 ms-1

(b) the magnitude of average velocity during time interval t = 0 to t = time of flight is 12 ms-1

(c) the magnitude of change in velocity from t = 0 to t = half of total time of flight is 9 ms-1

(d) the maximum height attain by the projectile is  4.05  m

Ans: (a) , (b) , (c) , (d)

Sol: Minimum Speed = uy = Vertical Component of 15 m/s

Minimum Speed = u sin37°

$\displaystyle = 15 \times \frac{3}{5} = 9 m/s$

Time of flight $\displaystyle T = \frac{2 u sin\theta}{g}$

$\displaystyle T = \frac{2 \times 15 \times sin37^o}{10}$

$\displaystyle T = \frac{2 \times 15 \times \frac{3}{5}}{10}$

$\displaystyle T = \frac{9}{5} sec$

Horizontal Range , $\displaystyle R = \frac{u^2 sin2\theta}{g}$

$\displaystyle R = \frac{15^2 (2 sin37^o cos37^o)}{10}$

$\displaystyle R = \frac{225 (2 \times \frac{3}{5}\times \frac{4}{5})}{10}$

R = 108/5 m

Average velocity in this duration $\displaystyle v_{avg} = \frac{108/5}{9/5} = 12 m/s$

After Half time of flight projectile is at highest point , where vertical component is zero .

Initial velocity $\displaystyle \vec{u} = u cos37^o \hat{i} + u sin37^o \hat{j}$

$\displaystyle \vec{u} = 15 (4/5) \hat{i} + 15 (3/5) \hat{j}$

$\displaystyle \vec{u} = 12 \hat{i} + 9 \hat{j}$

velocity after half time of flight , $\displaystyle \vec{v} = u cos37^o \hat{i}$ ; (vertical component is zero)

$\displaystyle \vec{v} = 15 (4/5) \hat{i}$

$\displaystyle \vec{v} = 12 \hat{i}$

Change in velocity $\displaystyle = 12 \hat{i} – (12 \hat{i} + 9 \hat{j})= -9 \hat{j}$

Maximum height attained , $\displaystyle H = \frac{u^2 sin^2 \theta}{2 g}$

$\displaystyle H = \frac{15^2 sin^2 37^o}{2 \times 10}$

$\displaystyle H = \frac{225 (3/5)^2}{20}$

$\displaystyle H = \frac{225 \times 9}{20 \times 25} = \frac{81}{10}$

H ≈ 4 m

## A particle is projected on smooth inclined plane in a direction perpendicular to line of greatest slope with 8 ms-1. Its speed at t= 1 s is

Q: A particle is projected on smooth inclined plane in a direction perpendicular to line of greatest slope with 8 ms-1. Its speed at t= 1 s is

(a) 5 ms-1

(b) 8 ms-1

(c) 6 ms-1

(d) 10 ms-1

Ans: (d)

## A projectile is projected on a smooth inclined plane of inclination 30°. The horizontal range is 120 m. …

Q: A projectile is projected on a smooth inclined plane of inclination 30°. The horizontal range is 120 m. If angle of projection is be 60°, the speed of projection is

(a) 37.2 ms-1

(b) 26.3 ms-1

(c) 20 ms

(d) 40 ms-1

Ans: (b)

Sol: The component of acceleration due to gravity along greatest slope is g’ = g sin30° = g/2 = 10/2 = 5 m/s2
angle of projection θ = 60°

Horizontal Range $\displaystyle R = \frac{u^2 sin2\theta}{g’}$

$\displaystyle 120 = \frac{u^2 sin2\times 60}{5}$

$\displaystyle u^2 = \frac{120\times 5 \times 2}{\sqrt{3}}$

$\displaystyle u = \sqrt{\frac{1200}{1.732}}$

u = 26.3 m/s

## A ball rolls from the edge of the top step of a stair case with horizontal speed 5 ms-1 . Each step is of 2 m high and 1 m width. The ball hits

Q: A ball rolls from the edge of the top step of a stair case with horizontal speed 5 ms-1 . Each step is of 2 m high and 1 m width. The ball hits

(a) 10th step

(b) 5th step

(c) 2nd step

(d) 3rd step

Ans: (a)

Sol: Let ball hits the nth steps & t be the time taken .

horizontal Distance , n × 1 = 5 t   …(i)

Vertical distance , $n \times 2 = \frac{1}{2}g t^2$  …(ii)

$2 n = \frac{1}{2}g (n/5)^2$   ; from (i)

$2 n = \frac{1}{2}(10) \times \frac{n^2}{25}$

n = 10