Question

A body is thrown with velocity (4i+3j)m/s. what is the horizontal range and time of flight of the projectile?

Answer 1

ello,

refer to the attachment please,

hope it helps you dear ❣️

Answer 2

Answer:

x component of initial velocity = 4m/s= ucosθ

y component of initial velocity = 3m/s = usinθ

Time of Flight = 2usinθ/g = 2*3/9.8 = 0.61s

Horizontal range = Time of Flight * velocity in x direction(x component)

                            = 0.61*4 = 2.44m

Please mark it as brainliest