This set of Mathematics 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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

_{0}∫

^{x}dx =

_{0}∫

^{x}(u – gt)dt

Or x = ut – (1/2)gt

^{2}……….(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,

_{u}∫

^{v}vdv = -g

_{0}∫

^{x}dx

Or v

^{2}= u

^{2}– 2gx ……….(5)

Let, t

_{1}be the time of rise of the particle; then v = 0, when t = t

_{1}.

Thus, from (2) we get,

0 = u – gt

_{1}

As, g = 9.8m/sec

^{2}

Or t

_{1}= 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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

_{0}∫

^{x}dx =

_{0}∫

^{x}(u – gt)dt

Or x = ut – (1/2)gt

^{2}……….(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)gT

^{2}

Or gT = 2u

Or T = 2u/g = 2(196)/9.8 [as, g = 9.8m/sec

^{2}]

= 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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

_{0}∫

^{x}dx =

_{0}∫

^{x}(u – gt)dt

Or x = ut – (1/2)gt

^{2}……….(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,

_{u}∫

^{v}vdv = -g

_{0}∫

^{x}dx

Or v

^{2}= u

^{2}– 2gx ……….(5)

If H m be the greatest height of the particle, then v = 0, when x = H; hence, from (5) we get,

0 = u

^{2}– 2gH

Or H = u

^{2}/2g

= (196*196)/(2*9.8) [as, g = 9.8 m/sec

^{2}]

= 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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

_{0}∫

^{x}dx =

_{0}∫

^{x}(u – gt)dt

Or x = ut – (1/2)gt

^{2}……….(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/sec

^{2}]

= 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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Since x = 0, when t = 0, hence from (2) we get,

_{0}∫

^{x}dx =

_{0}∫

^{x}(u – gt)dt

Or x = ut – (1/2)gt

^{2}……….(3)

Suppose the particle attains the height of 1254.4 m after t

_{2}seconds from the instant of projection.

Then, x = 1254.4, where t = t

_{2}; hence, from (3) we get,

1254.4 = 196t

_{2}– (1/2)(9.8)t

_{2}

^{2}[as, g = 9.8 m/sec

^{2}]

t

_{2}

^{2}– 40t

_{2}+ 256 = 0

Or (t

_{2}– 8)(t

_{2}– 32) = 0

Or t

_{2}= 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.

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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Let v

_{1}be the velocities of the particle after 10 seconds from the instant of projection.

Then v = v

_{1}when t = 10; hence from (2) we get,

v

_{1}= 196 – (9.8)*10 = 98 m/sec [as, g = 9.8m/sec

^{2}]

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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

Or v – u = -gt

Or dx/dt = u – gt ……….(2)

Let v

_{1}be the velocities of the particle after 10 seconds from the instant of projection.

Then v = v

_{1}when t = 10; hence from (2) we get,

v

_{1}= 196 – (9.8)*30 = -98 m/sec [as, g = 9.8m/sec

^{2}]

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

View Answer

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,

_{u}∫

^{v}dv = -g

_{0}∫

^{t}dt

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/sec

^{2}]

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**.

To practice all areas of Mathematics, __ here is complete set of 1000+ Multiple Choice Questions and Answers__.

**Next Steps:**

- Get Free Certificate of Merit in Mathematics - Class 12
- Participate in Mathematics - Class 12 Certification Contest
- Become a Top Ranker in Mathematics - Class 12
- Take Mathematics - Class 12 Tests
- Chapterwise Practice Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- Chapterwise Mock Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

**Related Posts:**

- Buy Class 12 - Mathematics Books
- Practice Class 12 - Physics MCQs
- Practice Class 12 - Chemistry MCQs
- Practice Class 12 - Biology MCQs
- Practice Class 11 - Mathematics MCQs