Question

A body is projected with a velocity of 10m/s at 45degrees to the horizontal. the velocity of the projectile when it moves at 30 degrees to the horizontal

Answer 1

As horizontal component of velocity remains same which is 5cos45°
And vertical velocity at starring is 5sin45°
Therefore when angle is 30° then velocity will be Vsin30°
By the given conditions
tan30° = Vsin30°/5cos45°
From solving above... We get V

Answer 2

Answer:

From projectile motion

R=u2sin2θg——(1)  

Here  R=  horizontal range of a projectile

u=20ms−1=  initial velocity of projection

g=9.8ms−2≈10ms−2=  Acceleration due to gravity

θ=450=  angle of projection

Therefore equation (1) becomes

R=20/2sin(90)10[sincesin90=1]  

⇒R=20×2010=20×2=40m