Question

A particle is projected with initial velocity u at an angle theta with the horizontal. Find the horizontal range travelled by the particle

Answer 1

Explanation:

here is your answer to the question

Answer 2

Answer:

let us consider the motion in x direstion

ux= uxCos theeta (by using vector resolution)

ax=0(there is no acceleration in x direction)

t=2u sin theeta/g

displacement (sx)= R

using : s=ut+1/2at square

R= ux cos theeta *2u sin theeta/g+0

R=u square 2 sin*cos theeta/g

      or

R=u sqr 2 sin theeta/g

Explanation: