Question

A stone is projected in air..Its time of flight is 3 sec. and range is 150 m..What is the horizontal component of velocity of projection of stone?​

Answer 1

Answer:

50 m/s

Explanation:

Range (R) of oblique projectile is calculated as

• R = uCosθ × T

where T is time of flight of projectile

150 = uCosθ × T

150 = uCosθ × 3

uCosθ = 150/3

uCosθ = 50 m/s

Answer 2

Answer:

The horizontal component of velocity of projection of stone is 100m/s

Explanation:

Given that,

A stone is projected in air with its time of flight being 3 sec and range is 150m.We are required to find the horizontal component of velocity.

First we write the given data in symbolic form as shown below

$T=3sec,R=150m$

If the angle of projection is $\theta$ and velocity of projection is u,then the relation for time of flight and range is

$T=\frac{2usin\theta}{g}\\R=\frac{u^{2}sin2\theta}{g}$

Substituting given values in above relation we get-

$3=\frac{2usin\theta}{g} = > usin\theta=\frac{3g}{2} -- > (1)$

Similarly $150=\frac{u^{2}sin2\theta}{g} = > (usin\theta)(ucos\theta)=150g--- > (2)$

Dividing equation 2 with equation 1,we get

$\frac{(usin\theta)(ucos\theta)}{usin\theta} =\frac{150g}{\frac{3g}{2} } \\= > ucos\theta=100m/s$

Therefore,The horizontal component of velocity of projection of stone is 100m/s.

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