A long jumper jumps at a 20-degree angle and attains a maximum altitude of 0.6 m. What is her initial speed? [10m/s] How far is
her jump? [6.59m]
I need someone to explain to me how to get the answers listed above my showing a detailed step to step process.
I need someone to explain to me how to get the answers listed above my showing a detailed step to step process.
1 answer:
Refer to the diagram shown below.
Assume g = 9.8 m/s² and ignore air resistance.
Let V = the initial speed
Let d = the distance raveled (how far she jumps).
The horizontal velocity is
Vx = V cos(20°) = 0.9397V m/s
The vertical launch velocity is
Vy = V sin(20°) = 0.342V m/s
At maximum height of 0.6 m, the vertical velocity is zero. Therefore
Vy² - 2*(9.8 m/s²)*(0.6 m) = 0
(0.342V m/s)² = 11.76 (m/s)²
0.342V = 3.4293
V = 10.027 m/s
Therefore
Vy = 0.342*10.027 = 3.4239 m/s
Vx = 0.9397*10.027 = 9.4226 m/s
The time to attain maximum height is a half of the total time of travel. This time is given by
Vy - gt = 0
3.4239 = 9.8*t
t = 3.4239/9.8 = 0.3494 s
The total travel time is 2*0.3494 = 0.6988 s
The length of the jump is
Vx*(2t) = (9.4226 m/s)*(0.6988 s) = 6.5845 m
Answer:
The initial speed is 10.0 m/s (nearest tenth)
The length of the jump is 6.58 m (nearest hundredth)
Assume g = 9.8 m/s² and ignore air resistance.
Let V = the initial speed
Let d = the distance raveled (how far she jumps).
The horizontal velocity is
Vx = V cos(20°) = 0.9397V m/s
The vertical launch velocity is
Vy = V sin(20°) = 0.342V m/s
At maximum height of 0.6 m, the vertical velocity is zero. Therefore
Vy² - 2*(9.8 m/s²)*(0.6 m) = 0
(0.342V m/s)² = 11.76 (m/s)²
0.342V = 3.4293
V = 10.027 m/s
Therefore
Vy = 0.342*10.027 = 3.4239 m/s
Vx = 0.9397*10.027 = 9.4226 m/s
The time to attain maximum height is a half of the total time of travel. This time is given by
Vy - gt = 0
3.4239 = 9.8*t
t = 3.4239/9.8 = 0.3494 s
The total travel time is 2*0.3494 = 0.6988 s
The length of the jump is
Vx*(2t) = (9.4226 m/s)*(0.6988 s) = 6.5845 m
Answer:
The initial speed is 10.0 m/s (nearest tenth)
The length of the jump is 6.58 m (nearest hundredth)
You might be interested in
Answer:
a)η = 69.18 %
b)W= 1210 J
c)P=3967.21 W
Explanation:
Given that
Q₁ = 1749 J
Q₂ = 539 J
From first law of thermodynamics
Q₁ = Q₂ +W
W=Work out put
Q₂=Heat rejected to the cold reservoir
Q₁ =heat absorb by hot reservoir
W= Q₁- Q₂
W= 1210 J
The efficiency given as
η = 69.18 %
We know that rate of work done is known as power
P=3967.21 W
Answer:
The volume of water displaced by the raft is 0.233 m³
Explanation:
The question relates to Archimedes' principle which states that the buoyant force experienced by an object immersed in a fluid is equal to the weight of (the force of gravity on) the displaced fluid
The given parameters are;
The combined mass of the person and the raft, m = 233 kg
The liquid on which the raft is located = Water
The density of water, = 1000 kg/m³
Weight = Mass, m × g
Where;
m = The mass of the object
g = The acceleration due to gravity = 9.8 m/s²
Given that the raft is on the surface of the water (floating), the buoyant force is equal to the combined weight of the person and the raft = 233 kg
The combined weight of the person and the raft, = 233 kg × 9.8 m/s² = 2,283.4 N
Therefore;
The buoyant force = 2,283.4 N = The weight of the water displaced
The mass of the water displaced, , = 2,283.4 N/(9.8 m/s²) = 233 kg
Density = Mass/Volume
The volume of water displaced by the raft = The mass of the water displaced/(The density of the water) = 233 kg/(1,000 kg/m³) = 0.233 m³.