## 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.

Click to See Solution :
Ans: 125/4 m
Sol: 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

Click to See Solution :
Ans: (A)
Sol: 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. 