# Class 12 Maths MCQ – Calculus Application – Motion Under Gravity

This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Motion Under Gravity”.

1. A particle is projected vertically upwards with a velocity of 196 m/sec. What will the time of rise?
a) 10 sec
b) 20 sec
c) 30 sec
d) 40 sec

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P, then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Since x = 0, when t = 0, hence from (2) we get,
0x dx = 0x (u – gt)dt
Or x = ut – (1/2)gt2 ……….(3)
Again, (1) can be written as,
dv/dx*dx/dt = -g
Or v(dv/dx) = -g ……….(4)
Since v = u, when x = 0, hence, from (4) we get,
uvvdv = -g 0xdx
Or v2 = u2 – 2gx ……….(5)
Let, t1 be the time of rise of the particle; then v = 0, when t = t1.
Thus, from (2) we get,
0 = u – gt1
As, g = 9.8m/sec2
Or t1 = u/g = 196/9.8 = 20 sec.

2. A particle is projected vertically upwards with a velocity of 196 m/sec. What is the value of total time of flight?
a) 40 sec
b) 45 sec
c) 50 sec
d) 55 sec

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Since x = 0, when t = 0, hence from (2) we get,
0x dx = 0x (u – gt)dt
Or x = ut – (1/2)gt2 ……….(3)
If T(≠0) be the total time of flight, then x = 0, when t = T; hence, from (3) we get,
Or 0 = uT – (1/2)gT2
Or gT = 2u
Or T = 2u/g = 2(196)/9.8   [as, g = 9.8m/sec2]
= 40 sec.

3. A particle is projected vertically upwards with a velocity of 196 m/sec. How much will be the greatest height?
a) 1930 m
b) 1960 m
c) 1990 m
d) 1995 m

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Since x = 0, when t = 0, hence from (2) we get,
0x dx = 0x (u – gt)dt
Or x = ut – (1/2)gt2 ……….(3)
Again, (1) can be written as,
dv/dx*dx/dt = -g
Or v(dv/dx) = -g ……….(4)
Since v = u, when x = 0, hence, from (4) we get,
uvvdv = -g 0xdx
Or v2 = u2 – 2gx ……….(5)
If H m be the greatest height of the particle, then v = 0, when x = H; hence, from (5) we get,
0 = u2 – 2gH
Or H = u2/2g
= (196*196)/(2*9.8)   [as, g = 9.8 m/sec2]
= 1960 m.

4. A particle is projected vertically upwards with a velocity of 196 m/sec. What will be its height from the point of projection after 12 sec?
a) 1646.2 m
b) 1645.4 m
c) 1644.2 m
d) 1646.4 m

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Since x = 0, when t = 0, hence from (2) we get,
0x dx = 0x (u – gt)dt
Or x = ut – (1/2)gt2 ……….(3)
Let, x = h when t = 12; then from (3) we get,
h = 196*12 – (1/2)*9.8*(12*12)  [as, g = 9.8m/sec2]
= 1646.4 m.

5. A particle is projected vertically upwards with a velocity of 196 m/sec. How many times will it attain a height of 1254.4 m after projection?
a) 0
b) 1
c) 2
d) 3

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Since x = 0, when t = 0, hence from (2) we get,
0x dx = 0x (u – gt)dt
Or x = ut – (1/2)gt2 ……….(3)
Suppose the particle attains the height of 1254.4 m after t2 seconds from the instant of projection.
Then, x = 1254.4, where t = t2; hence, from (3) we get,
1254.4 = 196t2 – (1/2)(9.8)t22  [as, g = 9.8 m/sec2]
t22 – 40t2 + 256 = 0
Or (t2 – 8)(t2 – 32) = 0
Or t2 = 8, 32
Clearly, the particle attains the height of 1254.4m twice; once after 8 seconds from the instant of its projection during its upward motion and next, after 32 seconds from the same instant during its motion in the downward direction.
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6. A particle is projected vertically upwards with a velocity of 196 m/sec. What will be its velocity at the end of 10 seconds?
a) 98 m/sec in the upward direction
b) 98 m/sec in the downward direction
c) 99 m/sec in the upward direction
d) 99 m/sec in the downward direction

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Let v1 be the velocities of the particle after 10 seconds from the instant of projection.
Then v = v1 when t = 10; hence from (2) we get,
v1 = 196 – (9.8)*10 = 98 m/sec   [as, g = 9.8m/sec2]
Since the upward direction is taken as positive, hence after 10 seconds the velocity of the particle is 98 m/sec in the upward direction.

7. A particle is projected vertically upwards with a velocity of 196 m/sec. What will be its velocity at the end of 30 seconds?
a) 98 m/sec in the upward direction
b) 98 m/sec in the downward direction
c) 99 m/sec in the upward direction
d) 99 m/sec in the downward direction

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Let v1 be the velocities of the particle after 10 seconds from the instant of projection.
Then v = v1 when t = 10; hence from (2) we get,
v1 = 196 – (9.8)*30 = -98 m/sec [as, g = 9.8m/sec2]
Since the upward direction is taken as positive, hence after 30 seconds the velocity of the particle is -98 m/sec in the upward direction. So, the velocity will be 98 m/sec in the downward direction.

8. A particle is projected vertically upwards with a velocity of 196 m/sec. When will its velocity be 49m/sec in the downward direction?
a) Before 23 sec
b) After 23 sec
c) Before 25 sec
d) After 25 sec

Explanation: Let the particle projected from A with velocity u = 196 m/sec and P be its position at time t where, AP = x m, if v m/sec be its velocity at P,then the equation of motion of the particle is,
dv/dt = -g [where, v = dx/dt] ……….(1)
Since v = u, when t = 0, hence from (1) we get,
uvdv = -g 0tdt
Or v – u = -gt
Or dx/dt = u – gt ……….(2)
Let the velocity of the particle be 49m/sec in the downward direction after T seconds from the instant of its projection. Then, v = -49 m/sec, when t = T; hence, from (2) we have,
-49 = 196 – (9.8)T  [as, g = 9.8 m/sec2]
Or 9.8(T) = 245
Or T = 25
Therefore, the particle will have a velocity of 49m/sec in the downward direction after 25 seconds from the instant of its projection.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.