A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.
The time of flight is
T = 2(Vo/g)sin28
The horizontal distance is
L = (Vo^2/g)sin56
The peak height is
H = Vo^2*sin^2(28)/(2g)
To find the time of flight, horizontal distance, and peak height of the long jumper, we need to break down the initial velocity into its horizontal and vertical components.
Given:
Initial velocity (v₀) = 12 m/s
Angle (θ) = 28 degrees above the horizontal
Step 1: Break down the initial velocity into horizontal and vertical components.
The horizontal component:
v₀x = v₀ * cosθ
v₀x = 12 m/s * cos(28°)
v₀x ≈ 10.61 m/s (rounded to two decimal places)
The vertical component:
v₀y = v₀ * sinθ
v₀y = 12 m/s * sin(28°)
v₀y ≈ 5.59 m/s (rounded to two decimal places)
Step 2: Find the time of flight.
The time of flight (t) can be calculated using the vertical component since the long jumper will be in the air when the vertical component is zero.
Using the equation:
v = v₀y + gt
Where:
v = final vertical velocity (when the jumper lands, v = 0)
g = acceleration due to gravity (which is approximately 9.8 m/s²)
0 = v₀y + gt
Rearranging the equation to solve for t:
t = -v₀y / g
t = (-5.59 m/s) / (-9.8 m/s²)
t ≈ 0.57 s (rounded to two decimal places)
Step 3: Calculate the horizontal distance.
The horizontal distance can be calculated using the horizontal component and the time of flight.
Using the equation:
d = v₀x * t
d = (10.61 m/s) * (0.57 s)
d ≈ 6.06 m (rounded to two decimal places)
Therefore, the time of flight is approximately 0.57 s, the horizontal distance is approximately 6.06 m, and the peak height is not yet determined.
To determine the time of flight, horizontal distance, and peak height of the long jumper, we will use the following kinematic equations:
1. Time of flight (t):
The time it takes for an object to reach the same vertical displacement during its ascent and descent is given by the equation:
t = 2 * (V * sinθ) / g
2. Horizontal distance (d):
The horizontal displacement of the object can be found using the equation:
d = V * cosθ * t
3. Peak height (h):
The maximum height reached by the object is given by the equation:
h = (V * sinθ)^2 / (2 * g)
Given:
Initial velocity (V) = 12 m/s
Angle (θ) = 28 degrees
Acceleration due to gravity (g) = 9.8 m/s^2
Let's calculate each value step-by-step:
1. Time of flight (t):
t = 2 * (V * sinθ) / g
t = 2 * (12 * sin(28)) / 9.8
t ≈ 1.22 seconds
2. Horizontal distance (d):
d = V * cosθ * t
d = 12 * cos(28) * 1.22
d ≈ 12.6 meters
3. Peak height (h):
h = (V * sinθ)^2 / (2 * g)
h = (12 * sin(28))^2 / (2 * 9.8)
h ≈ 2.18 meters
Therefore, the long jumper's time of flight is approximately 1.22 seconds, horizontal distance is approximately 12.6 meters, and the peak height is approximately 2.18 meters.