Initially (at time t = 0) a particle is moving vertically at 7.5 m/s and horizontally at 0 m/s. its horizontal acceleration is 1.4 m/s 2 . at what time will the particle be traveling at 49◦ with respect to the horizontal? the acceleration due to gravity is 9.8 m/s 2 .

t = 0.657 s

Explanation:

given,

initial vertical velocity = 7.5 m/s

initial horizontal velocity = 0 m/s

angle = 49◦

using kinetic equation

final velocity in vertical direction

v sinθ    = u_y   - gt ........................(1)

final velocity in horizontal direction

v cosθ = u_x   + a_x × t

here   u_x = 0.0 m/s

v cosθ = a_x×t ......................(2)

Dividing equation (1) / (2)

solving for time t

u_y   =   initial velocity along x direction

acceleration along a_x = 1.4 m/s²

g = acceleration due to gravity   = 9.8 m/s²

θ = 43° ,   u_y   =  7.5 m/s    

t = 0.657 s

time taken by the particle is t = 0.657 s

0.55x745.7j and a calculator

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this