a) To get how far from the base of the cliff does the ball land, we will use the SOH CAH TOA identity.
[tex]tan \theta =\frac{opposite}{adjacent}\\tan40^0=\frac{60}{x} \\x=\frac{60}{tan40^0}\\x=\frac{60}{0.8391}\\x= 71.5m[/tex]
This shows that the ball land 71.5m away from the base of the cliff.
b) The time the ball used in the air is the time of flight and it is expressed as:
[tex]T=\frac{2usin\theta}{g}\\T=\frac{2(90)sin40^0}{9.81}\\T=\frac{180(0.6428)}{9.81}\\T=\frac{115.704}{9.81}\\T= 11.794s[/tex]
Hence the time taken by the ball in the air is 11.79secs
c) The formula for calculating the maximum height is expressed as:
[tex]H=\frac{u^2sin^2 \theta}{2g} \\H=\frac{90^2sin^2(40)}{2(9.81)} \\H=\frac{8100(0.4132)}{19.62} \\H=\frac{3,346.92}{19.62} \\H=170.58m[/tex]
Hence the maximum height (above the cliff’s edge) that the cannonball attains is 170.58m.
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