A rugby player kicks a ball from the ground aiming to get the ball over the horizontal bar and between the vertical posts. The ball is 50m away from the goal line, directly in front of the goal, and the initial speed of the ball is 25m/s. The horizontal bar is 3.5 m from the ground. What are the two initial elevation angles between which the ball must be kicked to get it over the bar? Please explain your answer.

Thank you.

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Let alpha be the angle of inclination. Also let us assume that there is no resistance dus to air and the ball has negligible radius.

Then at any point, t

Distance travelled in the horizontal direction,

`x = ucosalpha*t` --- (i)

Distance travelled in the vertical direction,

`y = u sinalpha * t - 1/2gt^2` --- (ii)

From horizontal distance, we get

`t = x/ (u cosalpha)`

We substitute this in vertical equation to get

`y = u sinalpha * (x/ (u cosalpha)) - 1/2 * g * (x/ (u cosalpha))^2`

Upon simplifying,

`y = x * tanalpha-1/2 * g * x^2 * 1/(cos^2alpha)*1/u^2` --- (iii)

Again, `sin^2alpha +cos^2alpha = 1`

So,

`1/(cos^2alpha) = (sin^2alpha+ cos^2alpha)/ (cos^2alpha)`

`= tan^2alpha + 1`

We put this in equation (iii) to get

`y = x * tanalpha - 1/2*g *x^2 * ( 1 + tan^2alpha) / u^2`

Now, let `tanalpha =T`

`y = xT - 1/2 * g * x^2 * (1 + T^2)/u^2`

`y = 3.5, x = 50`

`rArr 3.5 = 50T - { 1/2 * (9.8) * (50 * 50) / (25 * 25) } * (1 + T^2)`

`rArr 3.5 = 50T - 19.6 - 19.6T^2`

`rArr 19.6T^2 - 50T + 23.1 = 0`

Solve quadratically to get,

`T = (50+-sqrt(50^2-4*19.6*23.1))/(2*19.6)`

`=1.945 and 0.606`

`tanalpha = 1.945 and 0.606`

`alpha = 62.8^o` and `31.21^o`

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