Question

A particle is projected with velocity 50m/s at an angle 60 degree with the horizontal from the ground. The time after which its velocity will make an angle 45 degree with the horizontal is

Answer 1

2 answers · Physics 

 Best Answer

To determine the angle of the velocity, we use the following equation. 
Tan θ = Vertical velocity ÷ Horizontal velocity 

As the particle moves from its initial position to its final position, its vertical velocity decreases at the rate of 9.8 m/s each second. During this time its horizontal velocity is constant. To determine the time when the angle is 45˚, we need to determine the object’s vertical velocity at this angle. Let’s determine the vertical and horizontal components of its initial velocity. 

Vertical = 50 * sin 60˚, Horizontal = 50 * cos 60˚ = 25 m/s 

Since horizontal velocity is constant, let’s put 25 m/s and 45˚ into the tangent equation. 

Tan 45 = Vertical velocity ÷ 25 
Vertical velocity = 25 m/s 
To determine the time for the vertical velocity to decrease from its initial value to 25 m/s, use the following equation. 

vf = vi – a * t, a = 9.8 
25 = 50 * sin 60˚ – 9.8 * t 
t = (25 – 50 * sin 60˚) ÷ -9.8 
This is approximately 1.87 seconds.

Answer 2

Answer:

Velocity along x axis(Ux)=u cos theta

=50cos60°

=50*root3/2

=25 root3m/s

Velocity along y axis(Uy)=u sin theta

=50sin60°

=50*1/2

=25m/s

Since, horizontal component remains constant,

Tan45°=Vy/Vx

1=Vy/25

Vy=25m/s

Vy=Uy+Ay*t

25=25 root3-10t [taking g=-10m/s^2]

t=25-25 root3/-10

t=25(1- root3)/-10

t=-25*0.732/-10

t=1.83s

Hope it will help you

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