A boy jumps at a speed of 16.0 m/s at an angle of 20.0 above the horizontal. How long does he stay in the air before touching the ground?
A.
1.53 s
B.
1.12 s
C.
3.07 s
D.
4.2 s
Vo = 16m/s[20o]
Yo = 16*sin20 = 5.47 m/s.
Y = Yo + g*t = 0 At max. Ht.
5.47 - 9.8t = 0
9.8t = 5.47
Tr = 0.558 s. = Rise time
Tf = Tr = 0.558 s. = Fall time.
T = Tr+Tf = 0.558 + 0.558 = 1.12 s.
To find the time the boy stays in the air before touching the ground, we can use the vertical component of his initial velocity and the acceleration due to gravity.
1. Split the initial velocity into its horizontal and vertical components:
- Vertical component: 16.0 m/s * sin(20.0°)
- Horizontal component: 16.0 m/s * cos(20.0°)
2. Determine the time it takes for the vertical component to reach 0 m/s using the equation of motion:
vf = vi + at
0 m/s = (16.0 m/s * sin(20.0°)) + (-9.8 m/s^2 * t)
Solve for t.
3. Solve for t:
-9.8 m/s^2 * t = -16.0 m/s * sin(20.0°)
t = (-16.0 m/s * sin(20.0°)) / -9.8 m/s^2
t ≈ 1.12 s
Therefore, the boy stays in the air for approximately 1.12 seconds before touching the ground.
The answer is B. 1.12 s.
To calculate the time the boy stays in the air before touching the ground, we can use the kinematic equation for vertical motion:
y = (v₀sinθ)t + (1/2)gt²
Where:
- y is the vertical displacement (in this case, the height is 0 because the boy touches the ground),
- v₀ is the initial vertical velocity (which can be calculated from the given speed and angle),
- θ is the launch angle (given as 20.0°),
- g is the acceleration due to gravity (approximately 9.8 m/s²),
- t is the time we want to find.
First, let's calculate the initial vertical velocity, v₀:
v₀ = v * sinθ
v₀ = 16.0 m/s * sin(20.0°)
v₀ = 16.0 m/s * 0.3420
v₀ ≈ 5.472 m/s
Now, we can substitute the known values into the equation and solve for t:
0 = (5.472 m/s)t + (1/2)(9.8 m/s²)t²
Simplifying the equation:
0 = 5.472t + 4.9t²
This is a quadratic equation which can be solved using the quadratic formula. The equation is in the form of ax² + bx + c = 0, where:
- a = 4.9 m/s² (coefficient of t²)
- b = 5.472 m/s (coefficient of t)
- c = 0 (constant term)
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Substituting the values:
t = (-5.472 ± √((5.472)² - 4(4.9)(0))) / (2(4.9))
Calculating:
t = (-5.472 ± √(29.942944 - 0)) / (9.8)
t = (-5.472 ± √(29.942944)) / (9.8)
t = (-5.472 ± 5.474) / (9.8)
This gives us two solutions:
- t = (-5.472 + 5.474) / (9.8) ≈ 0.002 s
- t = (-5.472 - 5.474) / (9.8) ≈ -1.1 s
Since time cannot be negative in this context, we discard the negative solution. So, the boy stays in the air for approximately 0.002 s.
The most appropriate answer from the given options is option B: 1.12 s.