Question

golf ball is projected at an angle of 30 degree with horizontal at a speed of 10 m/s. The angle made by line joining point of projection with the point of maximum height is

Answer 1

Concept:

• This projection will lead to the parabolic path of ball.
• at maximum height its y-component velocity = 0

Solution :

Given:

• initial velocity = u = 10 m/s
• horizontal velocity = $\rm u_x = ucos30\degree = 10\times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}$
• vertical velocity = $\rm u_y = usin30\degree = 10\times \dfrac{1}{2} = 5$
• $\alpha$ = ?

Time of flight of motion

$\implies\rm T = \dfrac{ 2u_y}{g} = \dfrac{2\times 5}{10} = 1s$

• Time taken at maximum height = T/2 = 1/2 = 0.5 s

At maximum height:-

$\implies \rm S_x = u_xt + \dfrac{1}{2}at^2$

$\to\rm X = 5\sqrt{3}(0.5)+ 0$

$\to\bf X = 2.5\sqrt{3} \: m$

$\implies \rm S_y = u_y t+ \dfrac{1}{2}at^2$

$\to\rm Y = 5(0.5)- \dfrac{1}{2}(10)(0.5)^2$

$\to\rm Y = 2.5 - 1.25$

$\to\bf Y = 1.25\: m$

Finding angle?

$\implies\rm tan\alpha = \dfrac{Y}{X}$

$\implies\rm tan\alpha = \dfrac{1.25}{2.5\sqrt{3}} = \dfrac{125\times 10}{25\sqrt{3}\times 100}$

$\implies\rm tan\alpha = \dfrac{5}{10\sqrt{3}} = \dfrac{1}{2\sqrt{3}}$

$\implies\bf \alpha = tan^{-1} \bigg( \dfrac{1}{2\sqrt{3}} \bigg)$

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Answer 2

Explanation:

Concept:

This projection will lead to the parabolic path of ball.

at maximum height its y-component velocity = 0

Solution :

Given:

initial velocity = u = 10 m/s

horizontal velocity = $\rm u_x = ucos30\degree = 10\times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}$

vertical velocity = $\rm u_y = usin30\degree = 10\times \dfrac{1}{2} = 5$

$\alpha$ = ?

Time of flight of motion

$\implies\rm T = \dfrac{ 2u_y}{g} = \dfrac{2\times 5}{10} = 1s$

• Time taken at maximum height = T/2 = 1/2 = 0.5 s

At maximum height:-

$\implies \rm S_x = u_xt + \dfrac{1}{2}at^2$

$\to\rm X = 5\sqrt{3}(0.5)+ 0$

$\to\bf X = 2.5\sqrt{3} \: m$

$\implies \rm S_y = u_y t+ \dfrac{1}{2}at^2$

$\to\rm Y = 5(0.5)- \dfrac{1}{2}(10)(0.5)^2$

$\to\rm Y = 2.5 - 1.25$

$\to\bf Y = 1.25\: m$

Finding angle?

$\implies\rm tan\alpha = \dfrac{Y}{X}$

$\implies\rm tan\alpha = \dfrac{1.25}{2.5\sqrt{3}} = \dfrac{125\times 10}{25\sqrt{3}\times 100}$

$\implies\rm tan\alpha = \dfrac{5}{10\sqrt{3}} = \dfrac{1}{2\sqrt{3}}$

$\implies\bf \alpha = tan^{-1} \bigg( \dfrac{1}{2\sqrt{3}} \bigg)$

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