The velocity of projection of a body is increased by 2%.Other factors remaining unchanged, what will be the percentage change in the maximum height attained?

Solution

From a point on the ground at a distance 2 meters from the foot of a vertical wall, a ball is thrown at an angle of 45º which just clears the top of the wall and afterward strikes the ground at a distance 4 m on the other side. The height of the wall is

²⁄₃m
³⁄₄m
¹⁄₃m
⁴⁄₃m

Solution

Three particles A, B and C are projected from the same point with same initial speeds making angles 30º, 45º and 60º respectively with the horizontal. Which of the following statements is correct?

A, B and C have unequal ranges
Ranges of A and C are equal and less than that of B
Ranges of A and C are equal and greater than that of B
A, B and C have equal ranges

Solution

A body is projected such that its KE at the top is 3/4 th of its initial KE. What is the angle of projectile with the horizontal?

30º
60º
45º
120º

Solution

If range is double the maximum height of a projectile, then θ is

tan^-14
tan^-11/4
tan^-1 1
tan^-1 2

Solution

\(\frac{u^{2}2sin\theta cos\theta }{g}=2\times \frac{u^{2}sin^{2}\theta }{2g}\, or\, tan\theta =2\)

A projectile thrown with a speed v at an angle θ has a range R on the surface of earth.For same v and θ, its range on the surface of moon will be\(\left \lfloor g_{moon}=\frac{g_{Earth}}{6} \right \rfloor\)

R/6
R
6 R
36 R

Solution

On earth, R= u²sin 2θ/g.

On moon, g'= g/6

R' = u²sin 2θ/g' = 6u²sin2θ/g = 6R.

Two balls are projected at an angle θ and (90º – θ) to the horizontal with the same speed. The ratio of their maximum vertical heights is

1 : 1
tanθ: 1
1 :tanθ
tan²θ: 1

Solution

\(\frac{H_{1}}{H_{2}}=\frac{u^{2}sin^{2}\theta }{u^{2}sin^{2}(90^{\circ}-\theta )/2g}=tan^{2}\theta\)

A gun fires two bullets at 60º and 30º with horizontal. The bullets strike at some horizontal distance. The ratio of maximum height for the two bullets is in the ratio

2 : 1
3 : 1
4 : 1
1 : 1

Solution

The range of a projectile when launched at an angle of 15º with the horizontal is 1.5 km. What is the range of the projectile when launched at an angle of 45º to the horizontal

1.5 km
3.0 km
6.3 km
0.75 km

Solution

A projectile is projected with a kinetic energy E. Its range is R.It will have the minimum kinetic energy, after covering a horizontal distance equal to

0.25 R
0.5 R
0.75 R
R

Solution

K.E. is minimum at the highest point. So, the horizontal distance is half of the range R i.e.,0.5 R.

UnAttempted

0

Attempted

0

Correct

0

Wrong

0

Complete