An athlete competing in the long jump leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal. How long does the athlete stay in the air, assuming no significant air resistance?
he is halfway through his jump when he reaches the top. That is when the vertical speed has gone to zero. That will be when
9.14 sin35° - 9.81t = 0
t = 9.14 sin35°/9.81
So he will be in the air for 2t seconds by the time he is back on the ground.
Well, if there's no significant air resistance, then there's no one to argue with the athlete on who gets to stay in the air longer. So, I'd say the athlete gets to stay in the air as long as they want! It's like a free vacation with no check-out time! Enjoy the high-flying adventure!
To determine how long the athlete stays in the air, we can use the equation of motion for the vertical direction. Let's break down the initial velocity into its vertical and horizontal components.
Given:
Initial speed (v₀) = 9.14 m/s
Angle (θ) = 35°
We can calculate the initial vertical velocity (v₀y) and the initial horizontal velocity (v₀x) using trigonometry:
v₀y = v₀ * sin(θ)
v₀y = 9.14 m/s * sin(35°)
v₀y ≈ 5.23 m/s
v₀x = v₀ * cos(θ)
v₀x = 9.14 m/s * cos(35°)
v₀x ≈ 7.47 m/s
Since there is no significant air resistance, the only force acting on the athlete in the vertical direction is gravity. The acceleration due to gravity (g) is approximately 9.8 m/s². Using this information, we can determine the time the athlete stays in the air.
The equation for vertical motion is:
y = v₀y * t - (1/2) * g * t²
Since the athlete starts and ends at the same elevation, the displacement in the vertical direction (y) is zero.
Setting y = 0, we can solve for the time (t):
0 = v₀y * t - (1/2) * g * t²
Rearranging the equation, we get:
(1/2) * g * t² = v₀y * t
Dividing both sides by t, we get:
(1/2) * g * t = v₀y
Now, solving for t:
t = (2 * v₀y) / g
Substituting v₀y = 5.23 m/s and g = 9.8 m/s² into the equation:
t = (2 * 5.23) / 9.8
t ≈ 1.05 seconds
Therefore, the athlete stays in the air for approximately 1.05 seconds, assuming no significant air resistance.
To determine the time the athlete stays in the air, we can use the equation of motion. In this case, we'll consider the vertical motion of the athlete.
The given information tells us that the athlete leaves the ground with a speed of 9.14 m/s at an angle of 35° above the horizontal. Since we want to know the time in the air, we only need to consider the vertical component of velocity.
The vertical component of the initial velocity can be calculated using the following equation:
Vy = V * sin(θ),
where Vy is the vertical component of velocity, V is the initial velocity, and θ is the angle of projection (35°).
Substituting the given values, we have:
Vy = 9.14 m/s * sin(35°),
Using a calculator, we find:
Vy ≈ 5.2 m/s.
Since the only force acting on the athlete during the vertical motion is gravity, we can use the equation of motion for vertical projectile motion:
Δy = Vy * t - (1/2) * g * t^2,
where Δy is the vertical displacement, Vy is the vertical component of velocity, t is the time, and g is the acceleration due to gravity (9.8 m/s²).
The vertical displacement, Δy, is zero because the athlete will return to the same vertical position as they started.
0 = (5.2 m/s) * t - (1/2) * (9.8 m/s²) * t^2.
Rearranging the equation, we get:
(1/2) * (9.8 m/s²) * t^2 - (5.2 m/s) * t = 0.
Factorizing the equation, we have:
t * [(1/2) * (9.8 m/s²) * t - (5.2 m/s)] = 0.
This equation gives us two solutions: t = 0 (which is not physically possible in this context), and the other solution is the time the athlete stays in the air:
(1/2) * (9.8 m/s²) * t - (5.2 m/s) = 0.
Simplifying the equation, we get:
4.9 m/s² * t = 5.2 m/s.
Finally, solving for t, we have:
t = 5.2 m/s / 4.9 m/s².
Calculating this expression, we find:
t ≈ 1.06 seconds.
Therefore, the athlete stays in the air for approximately 1.06 seconds.