## 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