Question

For projectile motion from ground to ground , horizontal and vertical component of initial velocities are 6 m/ s

and 8 m/s . Find Time of flight, Maximum height , Range and Angle of projection.​

Answer 1

Solution :-

As per the given data ,

• u cos θ = 6 m/s
• u sin θ = 8 m /s

Time of flight ,

➜ T = 2 x usinθ / g

➜ T = 2 x 8 / 10

➜ T = 1.6 s

The time of flight of the projectile is 1.6 s

Maximum height

➜ H = u² sin²θ / g

➜ H =( u sin θ )² / g

➜ H = 8 x 8 / 10

➜ H = 6.4 m

The maximum height of the projectile is 6.4 m

Range

➜ R = u² sin 2θ / g

➜R = u sinθ . u cosθ / g

➜ R = 8 x 6 / 10

➜R = 4.8 m

The range of the projectile is 4.8 m

Angle of projection

➜ tan θ = u sinθ / u cos θ

➜ tan θ = 8 / 6

➜ tan θ = 1.33

➜ θ = 53. 1

The angle of projection is 53. 1