Tags Projectile Motion

Projectile Motion

Get insights from 19 questions on Projectile Motion, answered by students, alumni, and experts. You may also ask and answer any question you like about Projectile Motion

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

a month ago

Projectile Motion Class 11th

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is y = 5x(1 – x). The y component vector of the initial velocity is ________ $\hat{j}$ .

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

y = x5 (1 – x) = x tan θ $(1 - \frac{x}{R})$

tan = 5, R = 1

$s i n θ = \frac{5}{\sqrt[]{26}}, c o s θ = \frac{1}{\sqrt[]{26}}$

$R = \frac{u^{2} s i n 2 θ}{g} = 1$

$\Rightarrow u^{2} = 26 \Rightarrow u = \sqrt[]{26} m / s$

y – component of initial velocity

= u sin θ

= $= \sqrt[]{26} * \frac{5}{\sqrt[]{26}}$

= 5 m/s

0 0

New answer posted

a month ago

Projectile Motion Class 11th

Two projectiles thrown at $30 ° a n d 45 °$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

H₁ = H₂

$\frac{u_{1}^{2} s i n^{2} θ_{1}}{2 g} = \frac{u_{2}^{2} s i n^{2} θ_{2}}{2 g}$

$\Rightarrow u_{1}^{2} {(s i n 30 °)}^{2} = {u_{2}}^{2} {(s i n 45 °)}^{2}$

$\Rightarrow$ $\frac{u_{1}}{u_{2}}$ $=2$

0 0

New answer posted

a month ago

Projectile Motion Class 11th

A monkey of mass 50kg climbs on a rope which can withstand the tension (T) of 350 N. If monkey initially climbs down with an acceleration of 4 m/s² and then climbs up with an acceleration of 5 m/s². Choose the correct option (g = 10 m/s²).

0 Follower 1 View

Payal Gupta

Contributor-Level 10

For climbing downward

50 g – T = 50a

T = 500 – 50 * 4

= 300 N < 350 N

for climbing upwards

T – 50 g = 50 a

T – 500 = 50 * 5

T = 750 N > 350 N

0 0

New answer posted

a month ago

Projectile Motion Class 11th

A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

$h = \frac{1}{2} g t^{2}$

or $x = v \sqrt[]{\frac{2 h}{g}}$

So, x = vt

or, $x = v \frac{2 h}{g}$

So, dist of man from helicopter is

$\sqrt[]{h^{2} + x^{2}} = \sqrt[]{h^{2} + v^{2} \frac{2 h}{g}} = \sqrt[]{\frac{2 v^{2} h}{g} + h^{2}}$

0 0

New answer posted

a month ago

Projectile Motion Class 11th

Given below are two statements. One is labeled as Assertion A and the other is labeled as Reason R.

Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then $R = 4 \sqrt[]{h_{1} h_{2}}$

Reason R : Product of said heights.

$h_{1} h_{2} = (\frac{u^{2} s i n^{2} θ}{2 g}) . (\frac{u^{2} c o s^{2} θ}{2 g})$

Choose the correct answer :

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

First angle, θ₁= θ, $R = \frac{u^{2} s i n 2 θ}{g}$

Another angle (θ₂ = 90 - θ) for which range will be Same as that of θ₁=θ

$R^{1} = \frac{u^{2} s i n 2 (90 - θ)}{g} = \frac{u^{2}}{g} s i n 2 θ = R$

at $θ_{1} = θ, h_{1} = \frac{u^{2} s i n^{2} θ}{2 g}$

$& θ_{2} = 90 - θ, h_{2} = \frac{u^{2} s i n^{2} θ_{2}}{2 g} = \frac{u^{2} c o s^{2} θ}{2 g}$

$\Rightarrow h_{1} h_{2} = \frac{1}{4^{2}} {[\frac{u^{2} s i n 2 θ}{g}]}^{2}$

$h_{1} h_{2} = \frac{1}{4^{2}} R^{2}$

$R = 4 \sqrt[]{h_{1} h_{2}}$

So, Both statement is true & Reason is correct explanation for statement 1.

0 0

New answer posted

2 months ago

Projectile Motion Motion in a Plane

A person moved from A to B on a circular path as shown in figure. If the distance travelled by him is 60m, then the magnitude of displacement would be:

(Given $c o s 135 °$ = -0.7)

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

d = R

$60 = R [\frac{3 π}{4}]$

$R = \frac{60 * 4}{3 π} = \frac{80}{π} m$

Displacement = $\sqrt[]{R^{2} + R^{2} - 2 R^{2} c o s 135}$ $^{}^{}^{}$

= $\sqrt[]{2 R^{2} - 2 R^{2} (- 0.7)}$ $^{}^{}$

= $\sqrt[]{3.4 R^{2}}$ $^{}$

= $\sqrt[]{3.4 {(\frac{80}{4})}^{2}}$ ${\frac{}{}}^{}$

47 m

0 0

New answer posted

2 months ago

Projectile Motion Class 11th

A ball is projected from the ground with a speed 15 ms^-1 at an angle θ with horizontal so that its range and maximum height are equal. Then 'tanθ' will be equal to

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$H = \frac{u^{2} s i n^{2} θ}{2 g} a n d R = \frac{u^{2} s i n 2 θ}{g}$

According to question

$R = H \Rightarrow \frac{u^{2} s i n^{2} θ}{2 g} = \frac{2 u^{2} s n θ c o s θ}{g} \Rightarrow t a n θ = 4$

0 0

New answer posted

2 months ago

Projectile Motion Class 11th

A ball of mass m is thrown vertically upwards. Another ball of mass 2 is thrown at an angle q with vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x} .$ The value of x is _______.

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$T_{1} = \frac{2 u}{g} T_{2} = \frac{2 V s i n θ}{g}$

$A s, T_{1} = T_{2} \Rightarrow \frac{2 u}{g} = \frac{2 V s i n θ}{g} \Rightarrow u = v s i n θ$ - (i)

$\frac{H_{1}}{H_{2}} = \frac{u^{2}}{2 g} * \frac{2 g}{v^{2} s i n^{2} θ} = {(\frac{u}{v s i n θ})}^{2} = 1$

0 0

New answer posted

2 months ago

Projectile Motion Class 11th

A body is projected from the ground at an angle of $45 °$ with horizontal. Its velocity after 2s is 20 ms^-1. The maximum height reached by the body during its motion is ______m. (use g = 10ms^-2)

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

after 2 sec, v = 20 m/sec

$\begin{array}{l} \vec{v} = u c o s 45 ° \hat{i} + (u s i n 45 ° - g t) \hat{j} \\ v = 20 = \sqrt[]{{(u c o s 45 °)}^{2} + {(u s i n 45 ° - 20)}^{2}} \end{array}$

$\Rightarrow 400 = u^{2} + 400 - 40 u s i n 45 °$

u = 40 sin 45°

$H_{m a x} = \frac{u^{2} s i n^{2} 45}{2 g} = \frac{4 0^{2} s i n^{2} 45 ° * s i n^{2} 45}{2 g}$

$= \frac{40 * 40}{2 * 10} * \frac{1}{2} * \frac{1}{2} = 20 m$

0 0

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
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Projectile Motion

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience