A 3-kg projectile is launhed at an angle of 45 o above the horizontal. The proje
ID: 1442985 • Letter: A
Question
A 3-kg projectile is launhed at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
A 3-kg projectile is launhed at an angle of 45o above the horizontal. The projectile explodes at the peak of its flight into two pieces. A 2-kg piece falls directly down and lands exactly 50 m from the launch point. Determine the horizontal distance from the launch point where the 1-kg piece lands.
75 m
200 m
100 m
150 m
75 m
200 m
100 m
150 m
Explanation / Answer
At the peak of its height the vertical velocity of the 3 kg projectile is 0 m/s. This occurs half way through the entire flight. Since the 2-kg piece falls directly down and lands exactly 50 m from the launch point, 50 meters is one half of the total horizontal distance. If the projectile did explode the total horizontal distance would be 100 meters. Let's use the range equation to determine the object's initial velocity.
Range = v^2/g * sin 2
100 = v^2/9.8
v = 980
This is approximately 31.3 m/s. Let's determine the initial vertical and horizontal velocity.
Vertical = 980 * sin 45, Horizontal = 980 * cos 45
Both of these velocities are approximately 22.14 m/s. As the object rises to the peak of its height, its vertical velocity decreases from approximately 22.14 m/s to 0 m/s at the rate of 9.8 m/s each second. The object's horizontal velocity is constant. At the peak of its height, its velocity is equal to the horizontal component its initial velocity. Let's determine the object's horizontal momentum at this position.
Momentum = 3 * 980 * cos 45
This is approximately 66.4.
Since the 2-kg piece falls directly down, its horizontal velocity at the maximum height is 0 m/s. During the explosion, its horizontal velocity decreased from approximately 22.14 m/s to 0 m/s. Let's determine the decrease of its horizontal momentum.
momentum = 2 * 980 * cos 45
This is approximately 44.27. To determine the horizontal momentum of the 1 kg object, subtract this from the total horizontal momentum.
M = 3 * 980 * cos 45 - 2 * 980 * cos 45 = 980 * cos 45
To determine the horizontal velocity of this piece, divide the momentum by its mass. Since its mass is 1 kg, its horizontal velocity is equal to the momentum.
v = 980 * cos 45
To determine the horizontal distance, we need to determine the time for it to fall from the maximum height to the ground. This time will be the same as the time for the object to rise to the peak of its height. Let's use the following equation to determine the time.
vf = vi - g * t, vf = 0, vi is the initial vertical velocity, 980 * sin 45
0 = 980 * sin 45 - 9.8 * t
t = 980 * sin 45 ÷ 9.8
The time is approximately 2.259 seconds. To determine the distance, multiply the horizontal velocity by the time.
d = 980 * cos 45 * 980 * sin 45 ÷ 9.8 = 980 * 0.5 ÷ 9.8 = 490/9.8 = 50 meters
The total horizontal is 100 meters.
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