Question

The horizontal range of a projectile is 4√3 times the maximum height achieved by it, then the angle of projection is?

Answer 1

Explanation:

R=4√3

so, u²sin2thita/g= 4√3.u²sin²thita/2g

2sinthitacosthita = 2√3sin²thita

costhita/sinthita = √3

cotthita = √3

thita = 30°

ans. 30°

Answer 2

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°