Four cannon balls, `(1),(2),(3), and (4)` are fired from level ground. Cannon ball (1) is fired at an angle of `60^@` above the horizontal and follows the path shown in (Fig. 5.9).
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Cannon balls (2) and (3) are fired at angle of `45^@` and (4) is fired at an angle of `30^@` above the horozontal. Which cannon ball has the largest initial speed ?

Explanation

Expressions for range and maximum height are given by
`R = (u^2 sin 2 theta)/(g) , H = (u^2 sin^2 theta)/(2 g)`
From (Fig. 5.9). (2) and (4) have same range.
`u_2^2 sin 2 xx 45^@ = u_4^2 sin 2 xx 30^@`
`u_2^2 sin^2 90^@ = u_4^2 sin^2 60^@`
Thus, `u_2^2 (1) = u_4^2 ((3)/(4))`
or `u_4 = (2)/(sqrt(3)) uor u_4 gt u_2`
Now we compare (2) and (3), `(Range)_3 lt (Range)_2`
As the projection angle is same, `u_3 lt u_2`.
Now we compare (1) and (2), the maximum height achieved by balls are same, `H_1 = H_2`.
or `(u_1^2 sin^2 60^@)/(2 g) = (u_2^2 sin^2 45^@)/(2 g) or u_1 = u_2 xx sqrt((2)/(3))`
or `u_2 gt u_1`
Thus, `u_1 gt u_2` and `u_2 gt u_1`
Therefore, `u_4` is maximum.
Hence, the initial velocity of ball (4) is maixmum.