(a) A girl threw a stone horizontally with a velocity of 30m/s from the top of a cliff 50m high. How far from the foot of the cliff does the stone strike the ground? [Take g= 10m/s\(^2\)
(b)
(b) A body A, of mass 2kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal.
Find the tensions T\(_1\), and T\(_2\), in the strings [Take g= 10ms\(^2\)]
(b)
(b) A body A, of mass 2kg is held in equilibrium by means of two strings AP and AR. AP is inclined at 56° to the upward vertical and AR is horizontal.
Find the tensions T\(_1\), and T\(_2\), in the strings [Take g= 10ms\(^2\)]
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Correct Answer: Option
Explanation:
(a) Time taken to reach the ground
Using S = ut + 1/2gt\(^2\)
(u= 0 for a body falling from rest)
50 = 0 x t + 1/2 x 10t\(^2\)
5t\(^2\) = 50;
t\(^2\) =50/5 = 10
t = √10 =3.16
Distance when it strikes the ground This journey is independent of gravity
d =vt =30 x 3.16 = 94.8m
(b) Using Lami's rule,
\(\frac{20}{sin 34}\) = \(\frac{T_1}{sin 90}\) = \(\frac{T_2}{sin 56}\)
= T\(_1\) \(\frac{20.sin 90}{sin 34}\) = T\(_2\) \(\frac{20.sin 56}{sin 34}\)
T\(_1\) = 35.79N and T\(_2\) = 29.65N
(a) Time taken to reach the ground
Using S = ut + 1/2gt\(^2\)
(u= 0 for a body falling from rest)
50 = 0 x t + 1/2 x 10t\(^2\)
5t\(^2\) = 50;
t\(^2\) =50/5 = 10
t = √10 =3.16
Distance when it strikes the ground This journey is independent of gravity
d =vt =30 x 3.16 = 94.8m
(b) Using Lami's rule,
\(\frac{20}{sin 34}\) = \(\frac{T_1}{sin 90}\) = \(\frac{T_2}{sin 56}\)
= T\(_1\) \(\frac{20.sin 90}{sin 34}\) = T\(_2\) \(\frac{20.sin 56}{sin 34}\)
T\(_1\) = 35.79N and T\(_2\) = 29.65N