5. The horizontal range of a projectile is 4v3 times of its maximum HU

1.	5. The horizontal range of a projectile is 4v3 times of its maximum HURUthe angle of projection.[0 = 306. A projectile is fired horizontally with a velocity of 98m/s from the top of a hill490m high. Find(a) The time taken to reach the ground(b) The distance of the target from the hill and(c) The velocity with which the projectile hits the ground.It10s, R=980m, va 138 59m/s
Answer» 5.ans=Answer: 30° is the angle of Projection Explanation: Let say Angle of Projection = α Horizontal Velocity = Vcosα Vertical Velocity = V sinα Velocity at max height = 0 using V = U + aT time = V sinα/g Max Height = V²sin²α/2g ( using V² - U² = 2aS) Time of Flight = 2V sinα/g Horizontal Distance = Vcosα * 2V sinα/g The horizontal range of a projectile is 4√3 times the maximum height achieved by it => Vcosα * 2V sinα/g = 4√3 V²sin²α/2g => 1/√3 = Tanα => α = 30° 30 is the angle of projectionanswer is 30° tan a=1/2 anda=30.

5. The horizontal range of a projectile is 4v3 times of its maximum HURUthe angle of projection.[0 = 306. A projectile is fired horizontally with a velocity of 98m/s from the top of a hill490m high. Find(a) The time taken to reach the ground(b) The distance of the target from the hill and(c) The velocity with which the projectile hits the ground.It10s, R=980m, va 138 59m/s

Answer»

5.ans=Answer:

30° is the angle of Projection

Explanation:

Let say Angle of Projection = α

Horizontal Velocity = Vcosα

Vertical Velocity = V sinα

Velocity at max height = 0

using V = U + aT time = V sinα/g

Max Height = V²sin²α/2g ( using V² - U² = 2aS)

Time of Flight = 2V sinα/g

Horizontal Distance = Vcosα * 2V sinα/g

The horizontal range of a projectile is 4√3 times the maximum height achieved by it

=> Vcosα * 2V sinα/g = 4√3 V²sin²α/2g

=> 1/√3 = Tanα

=> α = 30°

30 is the angle of projectionanswer is 30°

tan a=1/2 anda=30.