A long jumper leaves the ground at an angle of 20° from the horizontal at aspeed of
10.0 m/s. How long was the jumper in the air?
To determine how long the jumper was in the air, we can use the vertical component of the initial velocity and the acceleration due to gravity.
First, let's find the initial vertical velocity (Vy) using the initial speed (10.0 m/s) and the angle of takeoff (20°). We can use trigonometry to calculate this:
Vy = V * sin(θ)
Vy = 10.0 m/s * sin(20°)
Vy ≈ 3.42 m/s
Next, we can find the time it takes for the jumper to reach the peak of their jump. At the peak of the jump, the vertical velocity (Vy) will be zero. We can use the following kinematic equation:
Vy = V0y + gt
Where:
Vy is the final vertical velocity (which is 0 m/s at the peak)
V0y is the initial vertical velocity (3.42 m/s)
g is the acceleration due to gravity (-9.8 m/s²)
t is the time in seconds
0 = 3.42 m/s + (-9.8 m/s²) * t
9.8 m/s² * t = 3.42 m/s
t ≈ 0.35 s
Now, we can determine the total time of flight by doubling the time it takes to reach the peak of the jump. This is because the time it takes to reach the peak and the time it takes to descend are equal.
Total time of flight = 2 * 0.35 s
Total time of flight ≈ 0.70 s
Therefore, the long jumper was in the air for approximately 0.70 seconds.