Answer:
1) 20.49 ms⁻¹
2) ↓5 ms⁻¹ or -5 ms⁻¹
3) 300 ft/s
4)31.25 m
5) 17.32 ms⁻¹
Explanation:
1)v² = u² + 2as
v² = 0 + 2g(21)
v = 20.49 ms⁻¹
u = 0 as the object is dropped. let g = 10 ms⁻²
2)a = (v-u)/t ⇒↑ v = u + at (a = g)
= 30 - 10×3.5 = ↓5 ms⁻¹ or -5 ms⁻¹
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} = \frac{1}{2} mv^{2}\\v = u[/tex]
So velocity just before hitting ground is 300 ft/s
4) s = ut + (1/2)at²
= 0 + 0.5×10×(2.5)² = 31.25 m
5)v² = u² + 2as
v² = 0 + 2g(15)
v = 17.32 ms⁻¹
