Question

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 .

Answer 1

Answer:

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)

$tan \theta =\dfrac{u_y - gt}{a_x\times t}$

solving for time t

$t = \dfrac{u_y}{tan \theta \times a_x + g}$

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    

$t = \dfrac{7.5}{tan 49^0\times 1.4+ 9.8}$

t = 0.657 s

time taken by the particle is t = 0.657 s