Mathematics Questions and Answers – Calculus Application – Motion Under Gravity

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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

Answer: b
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.
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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

Answer: a
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
View Answer

Answer: b
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.
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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

Answer: a
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
View Answer

Answer: c
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
View Answer

Answer: a
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
View Answer

Answer: b
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.
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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

Answer: d
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.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter