SJIT ECE SOLITON PLACEMENT PREPARATION TEST@1 MARCH 24

A body of mass 5 kg, projected at an angle of 60° from the ground with an initial velocity of 25 m/s, acceleration due to gravity is g = 10 m/s², what is the maximum horizontal range covered?

• 49 m

• 54.13 m

• 49.16 m

• 60 m

A ball of mass 100 g, projected at an angle of 30° from the ground with an initial velocity of 11 m/s, acceleration due to gravity is g = 10 m/s², what is the maximum height attained?

• 1.5 m

• 3.0 m

• 1. 0 m

• 2.0 m

When do we get maximum range in a simple projectile motion?

• When θ = 45°

• When θ = 60°

• When θ = 90°

• When θ = 0°

When do we get maximum height in a simple projectile motion?

• When θ = 45°

• When θ = 90°

• When θ = 95°

• When θ = 75°

A football is projected at an angle of 45° from the ground with an initial velocity of 10 m/s, take acceleration due to gravity is g = 10 m/s².What is the time of flight?

• 1.4142 s

• 1.5361 s

• 1.8987 s

• 1.5651 s

A big stone of mass 1000 g, projected at an angle of 30° from the ground it covers a maximum vertical distance of 5 m, acceleration due to gravity is g = 10 m/s², what is the velocity with which it was thrown?

• 11.5 m/s

• 1.155 m/s

• 11.55 m/s

• 1.55 m/s

A soccer ball is projected at an angle of 60° from the ground.It attains its maximum height in 10s. Considering acceleration due to gravity as g = 10 m/s². What is the velocity with which it was projected?

• 11.55 m/s

• 120 m/s

• 117 m/s

• 115.5 m/s

A body of mass m, projected at an angle of θ from the ground with an initial velocity of v, acceleration due to gravity is g, what is the maximum horizontal range covered?

• R = v² (sin 2θ)/g

• R = v² (sin θ)/2g

• R = v² (sin 2θ)/2g

• R = v² (sin θ)/g