Question

A particle is projected from horizontal ground with speed 5ms^(-1) at 53^(@) with horizontal. Find time after which velocity of particle will be 45^(@) with horizontal:

Answer 1

Explanation:

a particle is projected Andrew reflects at a point so there answer is 5ms*(@).

Answer 2

The time after which the velocity of the particle will be 45° with the horizontal is 0.1 seconds.

Given,

Angle of projection = θ = 53°

Speed of projection = u = 5 m/s

To Find,

Time after which the velocity of the particle will be 45° with the horizontal

Solution,

We know that the 'x' component of projectile velocity always remains constant and in the horizontal direction.

uₓ = u Cos θ

uₓ = u Cos 53°

uₓ = 5 * $\frac{3}{5}$

uₓ = 3 m/s

The 'y' component of velocity keeps changing and is given as -

uy = u Sin θ

uy = u Sin 53°

uy = 5 * $\frac{4}{5}$

uy = 4 m/s

vy = uy - gt

where g is the acceleration due to gravity

and t is the time

vy = 4 - 10t   (Taking g = 10 m/s²)

According to the question,

$\frac{vy}{ux}$ = Tan 45°

$\frac{vy}{ux}$ = 1

vy = uₓ

vy = 3 m/s

vy = uy - gt

3 = 4 - 10t

10t = 1

t = $\frac{1}{10}$ seconds

t = 0.1 seconds

Thus, the time is 0.1 seconds.

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