Question

If an object is thrown at an angle 60°with respect to the horizontal with velocity 80m/sec. find time of flight.find horizontal range g=10m/sec^​

Answer 1

Solution:-

Range R . This is also equal to the displacement of the particles alnog x - axis in the t = T Thus applying

$\rm \: S = u_{x} t + \dfrac{1}{2} a_{x} {t}^{2}$

We get

$\rm \: R = (u \cos \alpha) \bigg( \dfrac{2u \sin\alpha }{g} \bigg) + 0$

as

$\rm \: a_{x} = 0 \: \: and \: \: t = T = \dfrac{2u \sin \alpha }{g}$

$\rm \: R = \dfrac{2 {u}^{2} \sin \alpha \cos \alpha }{g} = \dfrac{ {u^{2} \sin2 \alpha}}{g}$

$\boxed{\rm \: R \: = \dfrac{ {u^{2} \sin2 \alpha}}{g} }$

Given

$\rm \: u = 80m {s}^{ - 1}$

$\rm \: angle( \alpha) = 60 \degree$

$\rm \: g \: = 10ms {}^{ - 2}$

To find

$\rm \: horizontal \: range(R)$

Putting the value on formula

$\rm \: \to \: R = \dfrac{80 \times 80 \times \sin(2 \times 60) }{10}$

$\to \rm \: R = \dfrac{6400 \times \dfrac{ \sqrt{3} }{2} }{10}$

$\to \rm \: R = \dfrac{3200 \times \sqrt{3} }{10}$

$\rm \to \: R = 320 \times 1.732$

$\rm \to \: R = 554.24 \: m$

So horizontal range is 554 . 24 m

ii) Time of flight

=> Here x and y axis are in the direction shown in figure. axis x is along horizontal direction and axis y is vertical upward. Thus

$\rm \: u_{x} = u \cos \alpha$

$\rm \: u _{y } = u \sin \alpha \: \sf{and} \: \rm \: a _{x} = 0$

$\rm \: a _{y} = - g$

At point B , Sy = 0 . so applying

$\rm \: s_{y} = u_{y}t + \dfrac{1}{2} a _{y} {t}^{2}$

We have

$\rm \: 0 = (u \sin \alpha )t - \dfrac{1}{2} g {t}^{2}$

$\rm \: t = 0 \: \: and \: \: \dfrac{2u \sin \alpha }{g}$

So

$\boxed{ \rm \: T = \dfrac{2u \sin \alpha }{g} }$

putting the value on formula

$\rm \: T = \dfrac{2 \times 80 \times \sin60 \degree }{10}$

$\rm \: T = {16 \times \dfrac{ \sqrt{3} }{2} }{}$

$\rm \: T = 8 \sqrt{3} \: \: s$

$\rm \: T = 8 \times 1.732$

$\rm \: T = 13.86 \: \: s$

So time of flight 13.86 s

Answer 2

Explanation:

Range R . This is also equal to the displacement of the particles alnog x - axis in the t = T Thus applying

Step-by-step explanation:

ANSWER ✍️

Given

⇒Tanθ = 20/21

To Find

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

First of all We have to find Sinθ and Cosθ

So take

⇒Tanθ = 20/21 = Perpendicular(p)/Base(b)

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = h

Using Pythagoras theorem

⇒h² = p² + b²

⇒h² = (20)² + (21)²

⇒h² = 400 + 441

⇒h² = 841

⇒h = √(841)

⇒h = 29

We get

⇒Perpendicular = 20 , Base(b) = 21 and Hypotenuse(h) = 29

Then

⇒Sinθ = P/h and Cosθ = b/h

⇒Sinθ = 20/29 and Cosθ  = 21/29

Now Put the value on

⇒(1-Sinθ + Cosθ)/(1+Sinθ+Cosθ)

⇒(1-20/29 + 21/29)/(1+20/29 + 21/29)

⇒{(29-20+21)/29}/{29+20+21)/29}

⇒{(50 - 20)/29}/{(50+20)/29}

⇒(30/29)/(70/29)

⇒30/29 ×29/70

⇒30/70

⇒3/7

Answer = 3/7

hope this helps you!!

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