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.
Vx = 10 m/s
Vy = g*t = 9.8 m/s
Speed = sqrt[10^2 + 9.8^2] = ____
To solve this problem, we need to understand the concept of projectile motion. When a projectile is fired horizontally, it only has an initial horizontal velocity but no vertical velocity. However, gravity acts on the projectile, causing it to accelerate downward vertically.
When the projectile is fired horizontally, it moves with a constant horizontal velocity of 10 m/s. At the same time, gravity causes it to accelerate downward at a rate of 9.8 m/s².
Since there is no acceleration horizontally, the horizontal velocity remains constant throughout the entire motion. Therefore, after 1 second, the horizontal velocity is still 10 m/s.
However, vertically, the projectile undergoes a downward acceleration for 1 second. We can use the equation of motion to calculate its final vertical velocity (speed) after 1 second.
The equation for vertical motion without initial velocity is:
v = u + at
Where:
v = final velocity (speed)
u = initial velocity (speed) = 0 (since the projectile is fired horizontally)
a = acceleration due to gravity = 9.8 m/s²
t = time = 1 second
Plugging in the values:
v = 0 + (9.8 m/s²) * (1 s)
v = 9.8 m/s
Therefore, the speed of the projectile 1 second after it is fired is 9.8 m/s. However, this is only the vertical component of the velocity.
Since the horizontal and vertical components of velocity are independent of each other, the total speed of the projectile can be found using the Pythagorean theorem:
total speed = √(horizontal speed^2 + vertical speed^2)
total speed = √(10^2 + 9.8^2)
total speed ≈ 14 m/s
Therefore, the correct answer is B) 14 m/s.