Question

a particle is projected from ground at an angle of 60 above the horizontal with speed 150m/s. The time (in second) after which particle moves at an angle 45 with vertical,is

Answer 1

Answer:

Here in this case,

θ=45

o

,g=10m/s

2

, Horizontal distance (x)=10m,

Vertical distance (y)=7.5m

y=(tanθ)x −(

2u

2

cos

2

θ

g

)x

2

⇒7.5=(1×10)−

⎝

⎜

⎜

⎛

2u

2

×

2

1

10

⎠

⎟

⎟

⎞

×100

⇒

u

2

1000

=2.5

⇒u

2

=

2.5

1000

=400

⇒u=

400

=20 m/s.

Answer 2

Answer:

The time (in second) after which particle moves at an angle 45 with vertical,is 5.59sec

Explanation:

Given that,

A particle is projected from ground at an angle of 60° above the horizontal with speed 150m/s.

We are required to find the time after which the particle moves at an angle 45° with vertical.

Let the time taken by the particle to change the angle be t sec.

The horizontal velocity remains same throughout the motion.The horizontal velocity after t sec is

$V_{x}=u_{x}=150\cos60° = 150 \times \frac{1}{2} \\ = 75ms {}^{ - 1}$

The vertical velocity after t sec is

$V_{y}=150sin60°-9.81t \\ = 75 \sqrt{3} - 9.81t$

Given that inclination becomes 45° after time t.Therefore,

$\tan45°=\frac{V_{y}}{V_{x}}$

Substituting the values of the vertical and horizontal components in above equation,we get

$1 = \frac{75 \sqrt{3} - 9.81t }{75} \\ = > 75( \sqrt{3} - 1) = 9.81t \\ = > t = \frac{75 \times 0.732}{9.81} = 5.59 \: sec$

Therefore,The time (in second) after which particle moves at an angle 45 with vertical,is 5.59sec

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