Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
Horizontal component = (10N) · sin (20°) = 3.42... N (rounded)
Vertical component = (10N) · cos (20°) = 9.39... N (rounded)
The distance between two particles that are <em>in phase</em>
<u>Answers</u>
(a) 6.75 Joules.
(b) 5.27 m/s
(c) 0.75 Joules
<u>Explanation</u>
Kinetic energy is the energy possessed by a body in motion.
(a) its kinetic energy at A?
K.E = 1/2 mv²
= 1/2 × 0.54 × 5²
= 6.75 Joules.
(b) its speed at point B?
K.E = 1/2 mv²
7.5 = 1/2 × 0.54 × V²
V² = 7.5 ÷ 0.27
= 27.77778
V = √27.77778
= 5.27 m/s
(c) the total work done on the particle as it moves from A to B?
Work done = 7.5 - 6.75
= 0.75 Joules
