(a) A body of mass 15 kg is suspended at a point P by two
(a) A body of mass 15 kg is suspended at a point P by two light inextensible strings \(\overrightarrow{XP}\) and \(\overrightarrow{YP}\). The strings are inclined at 60° and 40° respectively to the downward vertical. Find, correct to two decimal places, the tensionsin the strings. [Take g = \(10 ms^{-2}\)].
(b) The height h metres, of a ball thrown into the air is \(2 + 20t + kt^{2}\), after t seconds. If it takes 2 seconds for the ball to reach its highest point, find
(i) the value of k (ii) its highest point from the point of throw.
Explanation
Using Lami's theorem, \(\frac{150}{\sin 80} = \frac{T_{1}}{\sin 40} = \frac{T_{2}}{\sin 60}\)
\(T_{1} = \frac{150 \sin 40}{\sin 80} = \frac{150 \times 0.6428}{0.9848} = 97.41 N\)
\(T_{2} = \frac{150 \sin 60}{\sin 80} = \frac{150 \times 0.8660}{0.9848} = 131. 90 N\)
(b) \(h = 2 + 20t + kt^{2}\)
\(\frac{\mathrm d h}{\mathrm d t} = v = 20 + 2kt\)
At the highest point, \(v = 0 ms^{-1}; t = 2 secs\)
\(\implies 20 + 2k(2) = 0\)
\(20 = -4k \implies k = -5\)
(ii) Its highest point
\(h = 2 + 20t - 5t^{2}\)
At highest point, t = 2 secs
\(h(2) = 2 + 20(2) - 5(2^{2}) = 2 + 40 - 20 = 22 m\)