Question

22. A particle is projected from ground with speed 80 m/
s at an angle 30° with horizontal from ground. The
magnitude of average velocity of particle in time
interval t = 2 s to t = 6 s is [Take g = 10 m/s21​

Answer 1

$\Huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Initial Velocity,u = 80 m/s
• Angle of Projection,∅ = 30°
• Time Interval,t = 2s to 6s

Time of Flight

$\huge{ \boxed{ \boxed{ \sf{t = \frac{2u \sin \theta}{g} }}}}$

Putting the values,we get:

$\sf{t = \frac{2 \times 80.sin30}{10} } \\ \\ \implies \: \huge{ \sf{t = 8s}}$

Since,

The heights attained by a projectile are equal at certain time intervals,the maximum height attained by the projectile is at 4s

For Instance,

The height of the projectile would be equal at 3s and 5s

Also,

• There is no vertical displacement

Thus,

Velocity along the Y - axis would be zero

Only velocity along X - axis comes into play : u.cos∅ = u.cos30

Now,

$\sf{ v = u.cos30} \\ \\ \implies \: \sf{v = 80. \frac{ \sqrt{3} }{2} } \\ \\ \implies \ \huge{\sf{v = 40 \sqrt{3} ms {}^{ - 1} }}$

Thus,the average velocity of the object is 40√3 m/s

Answer 2

♠️ $\huge\bold\green{HELLO}$♠️

GIVEN →

u = 80 m/s [INITIAL VELOCITY]

∅ = 30° [ANGLE OF PROJECTION]

t = 2s to 6s [TIME OF INTERVAL]

BY USING FORMULA , TIME OF FLIGHT →

$\huge\bold{T = \frac{</p><p>2usin \theta}{g}}</p><p>$

By substituting the values ,we get

$\sf{t = \frac{ t = 2 \times 80.sin30 }{10}}$

$\sf{t = \: 8 \: sec}$

So, The heights attained by a projectile are equal at certain time intervals,the maximum height attained by the projectile is at 4s

The height of the projectile would be equal at 3sec and 5sec

Also,There is no vertical displacement

So,Velocity along the Y - axis would be zero

Only velocity along X - axis comes into play : u.cos∅ {∅=30°} = u.cos30

So, now we find velocity by putting values

$\huge \sf{v = u.cos30} \\ \\ \sf{v = 80 \times \frac{ \sqrt{3} }{2} } \\ \\ \sf{v = 40 \sqrt{3}m {s}^{ - 1} }$