Question

please solve this question of maths you can solve this on page and attach your image of notebook​

Answer 1

Given :-

▪ A ball is thrown with initial velocity 50 m/s at an angle of projection 37°.

▪ Acceleration due to gravity, g = 10 m/s²

To Find :-

▪ Velocity vector and speed of the particle after 2 seconds of projection.

Solution :-

We know, the horizontal component of the velocity vector is constant throughout the projectile motion.

So, Horizontal component = u cos θ

⇒ horz. comp. = 50 × cos 37°

⇒ horz. comp. = 50 × 4/5

⇒ horz. comp. = 10 × 4

⇒ horz. comp. = 40 m/s

Now, Let us find the vertical component of the velocity vector at the beginning,

⇒ vert. comp. = u sin θ

⇒ vert. comp. = 50 × sin 37°

⇒ vert. comp. = 50 × 3/5

⇒ vert. comp. = 10 × 3

⇒ vert. comp. = 30 m/s

Now, Let us find the vertical component of the velocity vector after 2 seconds.

We have,

• Initial velocity, u = 30 m/s
• acceleration, a = g = - 10 m/s²
• Time, t = 2 seconds

Using first equation of motion,

⇒ v = u + at

⇒ v = 30 + (-10)2

⇒ v = 30 - 20

⇒ v = 10 m/s

Hence, The vertical component of the velocity vectors after 2 seconds is 10 m/s

Now, We have to find the speed which is the magnitude of the velocity vector.

⇒ Speed = √(10² + 40²)

⇒ Speed = √(100 + 1600)

⇒ Speed = √1700

⇒ Speed = 41.23 m/s

So, The Speed of the particle after 2 seconds will be 41.23 m/s

And, The velocity vector of the particle after 2 seconds will be 40i + 10j or 4i + j