Ask & Answer Question

ncert solutions physics class 11th physics ncert solutions class 11th Physics Class 11th Motion in a Plane

4.29 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

253 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

4.29 The angle of projectile $θ$ = 30 $°$

The bullet hits the ground at a distance of 3 km, Range R = 3 km

We know horizontal range for a projectile motion, R = $u^{2}$ sin2 $θ$ / g

3 = $u^{2}$ sin60 $°$ / g

$\frac{u^{2}}{g}$ = 3/ sin60 $°$ = 3.464 …………… (1)

To hit a target at 5 km,

Max Range, $R_{m a x}$ = $\frac{u^{2}}{g}$ …………. (2)

On comparing equation (1) and (2), we get

$R_{m a x}$ = 3.464

Hence, the bullet will not hit the target even by adjusting its angle of projection.

0 Upvotes 0 Downvotes

Similar Questions for you

A ball is projected from ground at angle q with the horizontal. After t = 1 s, it is moving at 45° with the horizontal and after t = 2 second, it is moving horizontally. What is speed of projection of ball? [g = 10 m s^–2]

alok kumar singh

Please find the solution below:

An electron in a research apparatus follows a circle path. On the electron’s first circuit of the apparatus, its speed is v₀ and the radius of its circular path is R. Each time it makes on circuit, it passes a short region where it receives a “kick” and gains an additional speed of $\frac{v_{0}}{100} .$ The electron follows a circular path such that the magnitude of its acceleration is always the same. What is the radius of the circular path after the electron has received 10 kicks?

Vishal Baghel

$a = \frac{v_{0}^{2}}{R}$ $\frac{_{}^{}}{}$

after 10 kicks, $v = v_{0} + \frac{10 v_{0}}{100} = 1.1 v_{0}$ $_{} \frac{_{}}{}_{}$

$\Rightarrow R_{1} = 1.21 R$

Two particles start moving from the same point in two different directions. The first particle moves along the positive x-axis with a constant acceleration of 3 m/s². The second particle moves in a straight line making an angle of 60° with the positive x-axis with a constant speed of 24 m/s. Find the time after which the relative velocity of the particles is minimum.

Vishal Baghel

v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:

Vishal Baghel

ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²

A flying disc of radius r is moving with constant speed v? while rotating anticlockwise with angular speed v?/r along a curve PQ as shown below. Radius of curvature of curve at the instant is 4r. The magnitude of acceleration of point A at the instant is (3v₀²)/(nr). Find n.

Vishal Baghel

a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

4.29 A bullet fired at an angle of 30° with the horizontal hits the ground 3.0 km away. By adjusting its angle of projection, can one hope to hit a target 5.0 km away? Assume the muzzle speed to be fixed, and neglect air resistance.

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question