Atop a cliff, a projectile is fired horizontally with an initial speed of 10 m/s. Neglecting air drag, what is its speed 1 second after it is fired? A) 20 m/s.B) 14 m/s.C) 30 m/s.D) 28 m/s.E) 10 m/s.
would this be 10 m/s?
No. See my previous answer.
http://www.jiskha.com/display.cgi?id=1330962875
No, the answer is not 10 m/s. The speed of the projectile 1 second after it is fired can be determined using the equations of motion. Since the projectile is fired horizontally, its initial vertical velocity is 0 m/s.
Using the equation for horizontal motion:
velocity (horizontal) = initial horizontal velocity = 10 m/s
The horizontal velocity remains constant throughout the motion.
Using the equation for vertical motion:
final vertical velocity = initial vertical velocity + (acceleration due to gravity * time)
Since the initial vertical velocity is 0 m/s, we can simplify the equation to:
final vertical velocity = acceleration due to gravity * time
Substituting the values:
final vertical velocity = (9.8 m/s^2) * (1 s)
final vertical velocity = 9.8 m/s
The speed of the projectile, which is the magnitude of its velocity, can be found using the Pythagorean theorem:
speed = sqrt[(velocity (horizontal))^2 + (final vertical velocity)^2]
speed = sqrt[(10 m/s)^2 + (9.8 m/s)^2]
speed = sqrt[100 m^2/s^2 + 96.04 m^2/s^2]
speed = sqrt[196.04 m^2/s^2]
speed = 14 m/s
Therefore, the correct answer is B) 14 m/s.
To solve this problem, we can use the concept of projectile motion.
When a projectile is fired horizontally, it only has an initial velocity in the horizontal direction and no initial velocity in the vertical direction. Neglecting air drag means that the horizontal velocity remains constant throughout the motion.
Given that the initial speed of the projectile is 10 m/s, we can determine the speed of the projectile 1 second after it is fired.
Since there is no initial vertical velocity, the only force acting in the vertical direction is gravity. The projectile will accelerate downwards, and the magnitude of this acceleration is approximately 9.8 m/s².
After 1 second, the vertical displacement of the projectile will be given by the formula: Δy = (1/2) * g * t², where g is the acceleration due to gravity (9.8 m/s²) and t is the time (1 second). Plugging in the values, we get: Δy = (1/2) * 9.8 * 1² = 4.9 meters.
Now, since the horizontal velocity remains constant, the speed of the projectile 1 second after it is fired will be the same as its initial speed of 10 m/s. Therefore, the correct answer is E) 10 m/s.