
Answer:
1) 20.49 msā»Ā¹
2) ā5 msā»Ā¹ or -5 msā»Ā¹
3) 300 ft/s
4)31.25 m
5) 17.32 msā»Ā¹
Explanation:
1)v² = u² + 2as
v² = 0 + 2g(21)
v = 20.49 msā»Ā¹
u = 0 as the object is dropped. let g = 10 msā»Ā²
2)a = (v-u)/t āā v = u + at Ā Ā (a = g)
Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā = 30 - 10Ć3.5 = ā5 msā»Ā¹ or -5 msā»Ā¹
3) By the conservation of Energy principle,
Energy of the arrow just after launch = energy of the arrow just before hit ground
[tex]\frac{1}{2} mu^{2} Ā = \frac{1}{2} mv^{2}\\v = u[/tex]
So velocity just before hitting ground is 300 ft/s
4) s = ut + (1/2)at²
Ā Ā Ā = 0 + 0.5Ć10Ć(2.5)² = Ā 31.25 m
5)v² = u² + 2as
v² = 0 + 2g(15)
Ā v = 17.32 msā»Ā¹