Problem Solving I A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.
A. A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
B. 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.
A. A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
B. 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.
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Answer:
A.
Given:
Initial Velocity (Vi) = 25 m/s
Angle of Release (θ) = 45 degrees
Acceleration Due to Gravity (g) = 9.8 m/s^2
To Find:
Time of Flight (t), Horizontal Distance (H), Peak Height (G)
Solution:
Using the given values and the equations of motion, we can find:
t = 2Vi sinθ / g
t = 2(25 m/s) sin 45 / 9.8 m/s^2
t = 3.21 seconds
H = Vi^2 sin2θ / g
H = (25 m/s)^2 sin(2 × 45)/ 9.8 m/s^2
H = 63.24 meters
G = Vi^2 sin^2θ / 2g
G = (25 m/s)^2 sin^2 45 / (2 × 9.8 m/s^2)
G = 31.62 meters
Therefore, the time of flight is 3.21 seconds, the horizontal distance is 63.24 meters, and the peak height is 31.62 meters.
B.
Given:
Initial Velocity (Vi) = 12 m/s
Angle of Release (θ) = 28 degrees
Acceleration Due to Gravity (g) = 9.8 m/s^2
To Find:
Time of Flight (t), Horizontal Distance (H), Peak Height (G)
Solution:
Using the given values and the equations of motion, we can find:
t = 2Vi sinθ / g
t = 2(12 m/s) sin 28 / 9.8 m/s^2
t = 1.67 seconds
H = Vi^2 sin2θ / g
H = (12 m/s)^2 sin(2 × 28) / 9.8 m/s^2
H = 7.21 meters
G = Vi^2 sin^2θ / 2g
G = (12 m/s)^2 sin^2 28 / (2 × 9.8 m/s^2)
G = 3.57 meters
Therefore, the time of flight is 1.67 seconds, the horizontal distance is 18.8 meters, and the peak height is 7.21 meters.
Explanation:
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