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 horizontal. Which cannon ball has the largest initial speed ?

Options

A. the initial velocity of ball (1) is maixmum
B. the initial velocity of ball (2) is maixmum
C. the initial velocity of ball (3) is maixmum
D. the initial velocity of ball (4) is maixmum

Answer

D

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.