Question

A body is projected at an angle 30° to the horizontal with a speed of 30 m/s. The angle made bythe velocity vector with the horizontal after 1.5 s is

Answer 1

Answer :-

→ 0° angle with horizontal .( vertical velocity will be 0 after 1.5 second )

To Find :-

→ Angle made by velocity vector with horizontal after 1.5 seconds .

Explanation :-

We have

• Projection angle = 30°
• Initial velocity (u) = 30 m/s

Hence ,

Vertical component of velocity is -

$\implies \sf{V_{v} = u \sin \theta \: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 30 \sin30 \degree = 15 \: \frac{m}{s}$

Horizontal component of velocity -

$\implies \sf{ V_{H} = u \cos \theta} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 30 \cos30 \degree = 15 \sqrt{3} \: \frac{m}{s}$

here no change in horizontal velocity because acceleration due to gravity works in vertical direction .

Now we have to find its velocity (v) after 1.5 second by using first equation of motion .

$\to \sf{v = V_{v}+ at\: \: \: \: \: \: \: \: \: ( a = - g) \: } \\ \\ \to \sf{ v = \sf{ 15 - g \times 1.5} }\\ \\ \to \sf{ v = 15 - 10 \times 1.5} \\ \\ \to \sf{v = 15 - 15 }\\ \\ \to \boxed{\sf{ v = 0 \: }}$

After 1.5 second its vertical velocity will ve zero means it makes 0° angle with horizontal.

→ Here total time of flight is 3 second means after 1.5 seconda it will at its maximum height where vertical velocity is zero .

Answer 2

$\tt{\huge{\orange {------------ }}} S.D$

QUESTION :

A body is projected at an angle 30° to the horizontal with a speed of 30 m/s.

The angle made by the velocity vector with the horizontal after 1.5 s is ...........

TO FIND :

The angle made by the velocity vector with the horizontal after 1.5 s ...

SOLUTION :

See the above attachment. It shows the F.B.D of the figure.

Now We first Calculate the horizontal and vertical components of velocity of the projectile..

Vx = V Cos Theta..

=> Vx = 30 × Cos ( 30° )

=> 30 × { √3 / 2 }

=> 15√{3} m / s.

Similarly we need to calculate the y component of velocity..

=> Vy = V × Sin Theta

=> Vy = 30 × sin ( 30 ° )

=> 30 × { 1 / 2 }

=> 15 m / s

We know that the first equation of motion :

V = u + at

Here , since we need to find the angle made bythe velocity vector with the horizontal after 1.5 s, we consider the X component of velocity.

Acceleration is constant, which is equal to -g.

=> V = 15 - 1.5 g

=> V = 15 ( 1 - g / 10 )

Since g Is approximately 10, the Resultant angle formed with the horizontal becomes 0 °

So, the angle formed with the vertical becomes 90°