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

Answer 2

The time at which the particle will be making 49° with respect to the horizontal is 0.657 sec.

Given to us

Initial vertical velocity = 7.5 m/s

Vertical acceleration = -g = -9.81 m/s²

Horizontal acceleration = 1.4 m/s²

θ = 49°

What is the final velocity in the vertical direction?

We know that according to the first equation of motion,

$v_y-u_y=a_y t$

$v_y-7.5=(-9.81) t\\\\v_y = 7.5-9.81 t\\\\v\ sin\theta = 7.5-9.81t$

What is the final velocity in the horizontal direction?

We know that according to the first equation of motion,

$v_x-u_x = at\\\\v_x = at\\\\v\ cos \theta =1.4t$

What is the final velocity of the particle?

To find the final velocity of the particle we will divide the final velocity in the vertical direction with the final velocity in the horizontal direction,

$\dfrac{v\ sin \theta}{v\ cos \theta} = \dfrac{7.5-9.81t}{1.4t}\\\\tan\theta = \dfrac{7.5}{1.4t} - \dfrac{9.81t}{1.4t}\\\\tan(49^o) + \dfrac{9.81t}{1.4t} = \dfrac{7.5}{1.4t}\\\\t = 0.657\rm\ sec$

Hence, the time at which the particle will be making 49° with respect to the horizontal is 0.657 sec.

Learn more about the First equation of motion:

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