Question

.

Answer 1

Part B :-


\(\large d = \frac{ v_0^2 }{ g }\sin(2\theta) \)

\(\large 7.68 = \frac{ v_0^2 }{ g }\sin(2\theta) \) ---------(1)

now, increase \(\large v_0\) by 3 %
so new initial velocity, \(\large \color{red}{V_0} = v_0 + 3 \% \ of \ v_0 \\ = v_0 + v_0 \times \frac{3}{100} \\ = v_0(1 + .03) \\ = 1.03 v_0 \)

So, wid this initial velocity, \(\large \color{Red}{V_0}\), the athlete wud go :-
\(\large D = \frac{ \color{red}{V_0}^2 }{ g }\sin(2\theta) \)
\(\large D = \frac{ (1.03v_0)^2 }{ g }\sin(2\theta) \) ------(2)

solve (1) and (2) for \\large D (\)

Answer 2

solve (1) and (2) for \( \large D \)

Answer 3

@pretty27