(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)