Projectile from a Height
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A stone is thrown in a vertically upward direction with a velocity of . If the acceleration of the stone during its motion is in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
An aeroplane is flying horizontally with a velocity of 600 km/h at a height of 1960 m. When it is vertically at a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is
1200 m
0.33 km
3.33 km
33 km
- 2h3
- 3h2
- 5h2
- 5h3
- 9 m/s
- 9√2 m/s
- 18 m/s
- 18√2 m/s
- 30∘
- 45∘
- 60∘
- 75∘
- particle B will reach at ground first with respect to particle A
- both particles will reach at ground simultaneously.
- both particles will reach at ground with same speed.
- particle A will reach at ground first with respect to particle B.
A bullet is dropped from the same height when another bullet is fired horizontally. They will hit the ground
One after the other
Simultaneously
Depends on the observer
None of the above
- North-west
- South-east
- tan−1(12) with east
- North
Match the quantities in column-I to their values in column-II
Column-I | Column-II |
(i) Speed after 2 s | p. 8√5 |
(ii)Time of flight | q. 12.8 |
(iii)Range | r. 115 |
(iv) The maximum height attained by the stone | s. 885 |
(v) Speed just before striking the ground. | t. √260 |
- (i) - p, (ii) - t, (iii) - s, (iv) - q, (v) - r
- (i) - t, (ii) - r, (iii) - s, (iv) - q, (v) - p
- (i) - p, (ii) - r, (iii) - s, (iv) - q, (v) - t
- (i) - r, (ii) - s, (iii) - p, (iv) - q, (v) - t
- 5 ms−1
- 10 ms−1
- 5 ms−1
- 20 ms−1
(i) The time taken to reach the ground
(ii)The distance of the target from the hill and
(iii)The velocity with which the projectile hits the ground. (take g=9.8 m/s2)
- (i) 5 s, (ii) 490 m, (iii) 110 m/s
- (i) 10 s, (ii) 980 m, (iii) 98√2 m/s
- (i) 15 s, (ii) 1470 m, (iii) 110 m/s
- (i) 5 s, (ii) 980 m, (iii) 98 m/s
- 3
- 9
- 2
- 4
- v2g
- √2v2g
- 2v2g
- √2[2v2g]
A particle is projected vertically upwards with a velocity of 20 m/sec. Find the time at which the distance travelled is twice the displacement
2+√43sec
1 sec
3 sec
2+√34
- Distance =3494 m, Velocity =√106 m/s
- Distance =√3492 m, Velocity =√106 m/s
- Distance =√3494 m, Velocity =106 m/s
- Distance =√3494 m, Velocity =√106 m/s
- 4003 m
- 5003 m
- 17003 m
- 498 m
- 0.5 ms−1
- 54 ms−1
- 5 ms−1
- 500 ms−1
- Slower one
- Faster one
- Both will reach simultaneously
- It cannot be predicted
- 2 km
- 0.2 km
- 20 km
- 4 km
(g=9.8 m/s2)
- 2 s
- √2 s
- 5√2 s
- 10√2 s
- 2 sec
- 4 sec
- 1 sec
- zero
(Take g=10 m/s2)
- tan−132
- tan−134
- 30∘
- 45∘
- 1 s
- 2 s
- 3 s
- 4 s
- 10√5 m/s
- 5√5 m/s
- 15 m/s
- 30 m/s
(Take g=10 m/s2)
- (a) 16 s (b) 5280 m (c) 366.7 m/s
- (a) 4 s (b) 1320 m (c) 330 m/s
- (a) 4 s (b) 1320 m (c) 332.4 m/s
- (a) 8 s (b) 2640 m (c) 332.4 m/s
- 660 m
- 1320 m
- 330 m
- 2640 m
- 2√2 sec
- 2 sec
- 4 sec
- 8 sec
An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth's surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling
On a parabolic path as seen by pilot in the plane
Vertically along a straight path as seen by an observer on the ground near the target
On a parabolic path as seen by an observer on the ground near the target
On a zig-zag path as seen by pilot in the plane
(Take g=10 m/s2)
- 5 ms−1
- 8 ms−1
- 10 ms−1
- 16 ms−1