Range
Trending Questions
Assertion: In javelin throw, the athlete throws the projectile at an angle of more than 45∘.
Reason: The maximum range does not depend upon the angle of projection.
Both assertion and reason are the correct answer, reason is the correct explanation for assertion.
Both assertion and reason are correct. The answer reason is the incorrect explanation for the assertion.
The assertion is correct and the reason is incorrect.
Both assertion and reason are incorrect.
- Speed before he jumps and his weight
- The direction in which he leaps and the initial speed
- The force with which he pushes the ground and his speed
- None of these
If the horizontal range of a projectile is equal to the maximum height reached, then the corresponding angle of projection is
- 22 m
- 6 m
- 15 m
- 11 m
Galileo writes that for angles of projection of a projectile at angles (45 + θ) and (45 - θ), the horizontal ranges described by the projectile are in the ratio of (if θ < 45)
2 : 1
1 : 2
1 : 1
2 : 3
- tan−1(14)
- tan−1(2)
- tan−1(4)
- π4
Distance between a frog and an insect on a horizontal plane is 9 m. Frog can jump with a maximum speed of √10 m/s . Minimum number of jumps required by the frog to catch the insect is (Take g=10 m/s2)
7
8
9
4
- 90∘
- 180∘
- 60∘
- 75∘
A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30∘ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground (g=10 m/s2, sin 30∘=12, cos 30∘=√32)
8.66 m
5.20 m
4.33 m
2.60 m
- 5√2 m/s
- 4√2 m/s
- 5 m/s
- 10√2 m/s
A cannon on a level plain is aimed at an angle α above the horizontal and a shell is fired with a muzzle velocity v0 towards a vertical cliff at a distance R away. Then the height from the bottom at which the shell strikes the side walls of the cliff is
R sin α−gR22v20sin2α
R cos α−gR22v20cos2α
R tan α−gR22v20cos2α
R tan α−gR22v20sin2α
- 4v25g
- 4g5v2
- v2g
- 4v2√5g
A ball is thrown from a point with a speed v0 at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v02 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection
Yes, 60∘
Yes, 30∘
No
Yes, 45∘
- 1:3
- 1:1
- 1:√3
- √3:1
A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed
100 m
150 m
200 m
250 m
- 1:3
- 1:1
- 1:√3
- √3:1
- Their ranges are same
- Their heights are same
- Their times of flight are same
- All of these
A bullet is fired from a cannon with velocity 500 m/s. If the angle of projection is 15∘ and g=10m/s2. Then the range is
25×103m
12.5×103m
50×102m
25×102m
- 22 m
- 6 m
- 15 m
- 11 m
An aeroplane moving horizontally with a speed of 720 km/h drops a food pocket, while flying at a height of 396.9 m. the time taken by a food pocket to reach the ground and its horizontal range is (Take g = 9.8 m/sec2)
3 sec and 2000 m
5 sec and 500 m
8 sec and 1500 m
9 sec and 1800 m