Answer:
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Answer:
Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point.
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Answer:
<h2>B. 20°</h2>
Explanation:
Range in projectile is defined as the distance covered in the horizontal direction. It is expressed as R = U²sin2Ф/g
U is the initial velocity of the body (in m/s)
Ф is the angle of projection
g is the acceleration due to gravity.
Given U = 14m/s, g = 9.8m/s and range R = 15 m
we will substitute this value into the formula to get the projection angle Ф as shown;
15 = 15²sin2Ф/9.8
15*9.8 = 15²sin2Ф
147 = 225sin2Ф
sin2Ф = 147/225
sin2Ф = 0.6533
2Ф = sin⁻¹0.6533
2Ф = 40.79°
Ф = 40.79°/2
Ф = 20.39° ≈ 20°
Hence, the range is greatest at angle 20°
Answer:
The horizontal displacement of the arrow is not larger than the banana split.
Explanation:
Using y - y₀ = ut - 1/2gt², we find the time it takes the arrow to drop to the ground from the top of mount Everest.
So, y₀ = elevation of Mount Everest = 29029 ft = 29029 × 1ft = 29029 × 0.3048 m = 8848.04 m, y = final position of arrow = 0 m, u = initial vertical speed of arrow = 0 m/s, g = acceleration due to gravity = 9.8 m/s² and t = time taken for arrow to fall to the ground.
y - y₀ = ut - 1/2gt²
0 - y₀ = 0 × t - 1/2gt²
-y₀ = -1/2gt²
t² = 2y₀/g
t = √(2y₀/g)
Substituting the values of the variables, we have
t = √(2y₀/g)
= √(2 × 8848.04 m/9.8 m/s²)
= √(17696.08 m/9.8 m/s²)
= √(1805.72 s²)
= 42.5 s
The horizontal distance the arrow moves is thus d = vt where v = maximum firing speed of arrow = 100 m/s and t = 42.5 s
So, d = vt
= 100 m/s × 42.5 s
= 4250 m
= 4.25 km
Since d = 4.25 km < 7.32 km, the horizontal displacement of the arrow is not larger than the banana split.
