# Projectile Time, Height and Range

## Trending Questions

**Q.**

For a projectile, the ratio of maximum height reached to the square of flight time is g=10ms−2

**Q.**

How do you evaluate $sec120\xc2\xb0$?

**Q.**

For a body to escape from earth, angle at which it should be fired is

${45}^{0}$

$>{45}^{0}$

$<{45}^{0}$ and

any angle

**Q.**A body is projected at 30∘ with velocity v1 from point A on the ground. At the same time another body is projected vertically upwards from B with velocity v2 from the ground. The point B lies vertically below the highest point of the trajectory of ball A. For both the bodies to collide, what should be the ratio of v1/v2 ?

- 1
- √2
- 12
- 2

**Q.**A stone is projected from the point on the ground in such a direction so as to hit a bird on the top of a telegraph post of height h and then attain the maximum height 3h2 above the ground. If at the instant of projection, the bird were to fly away horizontally with uniform speed, find the ratio between horizontal velocities of the bird and stone if the stone still hits the bird while descending.

- 1:1
- 1:2
- 1:√3+1
- √3−1:1

**Q.**A jet of water is projected at an angle θ=45∘ with the horizontal from a point which is at a distance of x=15 m from a vertical wall as shown in the figure. If the speed of projection is 10√2 m/s, find out the height from ground at which the water jet strikes vertical wall. Take g=10 m/s2

- 5 m
- 3.75 m
- 7.5 m
- 2.5 m

**Q.**The distance between a frog and an insect is 12 m on a horizontal surface. The frog can jump with a maximum speed of √10 m/s. The minimum number of jumps required by the frog to catch the insect is

[ Take g=10 m/s2 ]

- 13
- 10
- 11
- 12

**Q.**A projectile can have the same range for two angles of projection. If h1 and h2 are maximum heights when the range in the two cases is R, then the relation between R, h1 and h2 is

- R=4√h1h2
- R=2√h1h2
- R=√h1h2
- None of these

**Q.**A particle is projected at an angle of 30∘ w.r.t horizontal with speed 20 m/s. Take the point of projection as the origin of the coordinate axes and find the angle between velocity vector and position at t=1 s

- cos−1(√313)
- cos−1(2√313)
- cos−1(4√313)
- 30∘

**Q.**The maximum height reached by the projectile is 4 m. The horizontal range is 12 m. Velocity of projection (in ms−1) is

(g is acceleration due to gravity)

- 5√g2
- 3√g2
- 13√g2
- 15√g2

**Q.**A projectile is thrown at an angle of 60∘ with horizontal. If E is the kinetic energy with which the projectile is thrown, then find the kinetic energy at the top of trajectory.

- 3E4
- E4
- E2
- √3E2

**Q.**A particle is projected at an angle with the horizontal such that it follows a trajectory given by the equation y=6x–2x2. Find the maximum height attained by it.

- 2.5 m
- 5.2 m
- 4.5 m
- 6 m

**Q.**The horizontal range and maximum height attained by a projectile are R and H, respectively. If a constant horizontal acceleration a=g4 is imparted to the projectile due to wind, then its horizontal range and maximum height will be

- (R+H), H2
- (R+H), 2H
- (R+2H), H
- (R+H), H

**Q.**Two buildings A and B are 20 m apart. Building A has a window at a height of 50 m above the ground, and building B has a window at height 30 m from the ground. At what speed must a man throw a ball horizontally from the window of building A so that it enters the window of building B. Take g=10 m/s2.

- 5 m/s
- 15 m/s
- 10 m/s
- 20 m/s

**Q.**A particle is projected at an angle of 30∘ w.r.t horizontal with speed 20 m/s. Take the point of projection as the origin of the coordinate axes and find the velocity vector of the particle after 1 s (in m/s).

- 5√3^i+5^j
- 5^j
- 5^i+10√3^j
- 10√3^i

**Q.**A particle is projected at an angle of 30∘ w.r.t horizontal with speed 20 m/s. Take the point of projection as the origin of the coordinate axes and find the position vector(in m) of the particle after 1 s. (Take g=10 m/s2)

- 5√3^i+5^j
- 10√3^i+5^j
- 5^i+10√3^j
- 10√3^i+10^j

**Q.**A projectile is thrown with a speed of 100 m/s making an angle of 60∘ with the horizontal. Find the minimum time after which its inclination with the horizontal is 45∘.

- 5(√3−1)
- 5(√2−1)
- 5(1−√3)
- 5(1−√2)

**Q.**A projectile motion is projected with a velocity μ at an angle θ, with the horizontal. For a fixed θ, which of the graphs shown in the following figure shows the variation of R versus μ

**Q.**A ball is rolled off the edge of a horizontal table at a speed of 4 m/s. It hits the ground after 0.4 second. Which statement given below is true(Take g=10m/s2)

- It hits the ground at a horizontal distance 1.6 m from the edge of the table
- The speed with which it hits the ground is 4.0 m/second
- Height of the table is 1.8 m
- It hits the ground at an angle of 60
^{o}to the horizontal

**Q.**Two particles A and B initially separated by distance X are projected simultaneously in the directions shown in figure with speed vA=20 m/s and vB=10 m/s respectively. They collide in air after 12 s. Find the distance X.

- 10√5 m
- 5√3 m
- 4√6 m
- 5√2 m

**Q.**In the given figure for a projectile

Which of the following option is correct?

- y=[x1x2x1−x2]tan θ
- y=[x1x2x1+x2]tan θ
- y=[2x1x2x1+x2]cos θ
- y=[2x1x2x1+x2]tan θ

**Q.**

A projectile has an initial velocity ${v}_{0}$ at an angle $\mathrm{\xce\xb8}$ above the horizontal. It reaches the highest point of its trajectory in time $T$ after the launch. The highest point is at a vertical distance $H$ and at the horizontal distance $d$ from the point of projection. The speed of projectile at its highest point is $v$. For this situation, mark out the correct statement(s).

$d={v}_{0}\mathrm{cos}\mathrm{\xce\xb8}\xc3\u2014T$

$v={v}_{0}\mathrm{cos}\mathrm{\xce\xb8}$

$H=\frac{{\left({v}_{0}\mathrm{sin}\mathrm{\xce\xb8}\right)}^{2}}{2g}$

$H=\frac{g{T}^{2}}{2}$

**Q.**The range of projectile is R when the angle of projection is 55∘. For the same velocity of projection and range, the other possible angle of projection is

- 60∘
- 45∘
- 35∘
- Not possible

**Q.**The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

- tan−1(14)
- tan−1(2)
- tan−1(4)
- π4

**Q.**A particle of mass m is projected from the ground with initial linear momentum p (magnitude) so as to have maximum possible range. Its minimum kinetic energy will be

- p22m
- p24m
- p2m
- p26m

**Q.**A particle is projected with a velocity u at two different angles which are complementary to each other. If H1 and H2 are the heights attained by them in each case, find the range of the particle.

- 4√H1H2
- 2√H1H2
- √H1H2
- 12√H1H2

**Q.**The velocity at the maximum height of a projectile is √32 times its initial velocity of projection (u). Its range on the horizontal plane is

- √3u22g
- 3u22g
- 3u2g
- u22g

**Q.**For a projectile thrown with a speed v, the horizontal range is √3v22g. The vertical range is v28g. The angle which the projectile makes with the horizontal initially is

- 15∘
- 30∘
- 45∘
- 60∘

**Q.**A projectile is fired at an angle of 45∘ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

- tan−1(√32)
- 45∘
- 60∘
- tan−112

**Q.**A particle is projected from the ground with velocity u at an angle θ with horizontal. The horizontal range, maximum height and time of flight are R, H and T respectively. They are given by, R=u2sin2θg, H=u2sin2θ2g and T=2usinθg

Now keeping u as fixed, θ is varied from 30∘ to 60∘. Then

- R will first increase then decrease, H will increase and T will decrease
- R will first increase then decrease while H and T both will increase
- R will decrease while H and T will increase
- R will increase while H and T will increase