15a. A body of mass 15kg is suspended at a point P by two light inextensible strings XP\(^→\) and YP\(^→\). The strings are inclined at 60º and 40º, respect...
Question 1 Report
15a. A body of mass 15kg is suspended at a point P by two light inextensible strings XP\(^→\) and YP\(^→\). The strings are inclined at 60º and 40º, respectively, to the downward vertical. Find, correct to two decimal places, the tension in the strings (take g = 10m/s\(^2\))
b. The height h metres, of a ball thrown into the air is 2 + 20t + kt\(^2\), after t seconds. If its takes 2 seconds for the ball to reach its height point, Find:
i. the value of k
ii. its highest point from the point of throw.
Answer Details
\(\frac{150}{sin 100} = \frac{T_1}{sin 140}\)
T\(_1\) = \(\frac{150 sin 140}{sin 100}\) = 97.91N
In same manner:
\(\frac{150}{sin 100} = \frac{T_2}{sin 120}\)
T\(_2\) = \(\frac{150 sin 120}{sin 100}\)
T\(_2\) = 131.91N.
i. At the highest point, v = 0 when t = 2 s:
\(20 + 2k \cdot 2 = 0 \)
20 + 4k = 0
k = - 5
ii. The highest point "from the point of throw" is the upward displacement above the release point t = 0, where the variable terms begin.
At t = 2:
2 + 20 x 2 + k x (2)\(^2\) = 2 + 40 + (-5) x 4 = 42 - 20 = 22 m.
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