A track star in the broad jump goes into the jump at 10 m/s and launches himself at 20° above the horizontal. How long is he in the air before returning to Earth? Note: g = 9.8 m/s2.Select one:a.0.70 sb.0.89 sc.0.42 sd.0.35 s
Question
A track star in the broad jump goes into the jump at 10 m/s and launches himself at 20° above the horizontal. How long is he in the air before returning to Earth? Note: g = 9.8 m/s2.Select one:a.0.70 sb.0.89 sc.0.42 sd.0.35 s
Solution 1
To solve this problem, we need to use the equations of motion. The time of flight in projectile motion can be calculated using the formula:
t = 2u sin θ / g
where: u = initial velocity = 10 m/s θ = angle of projection = 20° g = acceleration due to gravity = 9.8 m/s²
First, we need to convert the angle from degrees to radians because the trigonometric functions in calculators use radians.
θ = 20° = 20 * π/180 = 0.349 radians
Now, we can substitute the values into the formula:
t = 2 * 10 * sin(0.349) / 9.8 t = 1.176 seconds
However, this answer is not in the options. There might be a mistake in the question or the options.
Solution 2
To solve this problem, we need to use the equations of motion. The time of flight in projectile motion can be calculated using the formula:
t = 2u sinθ / g
where: u = initial velocity = 10 m/s θ = angle of projection = 20° g = acceleration due to gravity = 9.8 m/s²
First, we need to convert the angle from degrees to radians because the trigonometric functions in calculators use radians.
θ = 20° = 20 * π/180 = 0.349 radians
Now, we can substitute the values into the formula:
t = 2 * 10 * sin(0.349) / 9.8 t = 1.176 seconds
However, this answer is not in the options. It seems there might be a mistake in the question or the options provided.
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