Answer :
Final answer:
The speed of a projectile 3 seconds after launch at an angle remains the same as its initial speed, 120 m/s, in the absence of air resistance. This is because horizontal velocity is constant and the change in vertical velocity due to gravity does not affect the overall magnitude of the speed of the projectile. The correct answer is D.
Explanation:
The student asked about the speed of a projectile motion 3.00 seconds after it is launched at 35.0° above the horizontal with an initial velocity of 120 m/s. In the absence of air resistance, the speed of a projectile does not change in horizontal motion, and the only acceleration acting on the projectile is gravity, which affects the vertical component of the motion. Thus, the horizontal velocity remains constant, and the vertical velocity changes due to gravity (9.8 m/s² downward).
To find the projectile's speed after 3 seconds, we need to calculate the vertical velocity at that time using the formula Vy = u + at, where u is the initial vertical velocity, a is the acceleration due to gravity, and t is the time. Then we find the resultant speed using the horizontal velocity (which remains 120 m/s × cos(35.0°)) and the new vertical velocity.
Calculation of Vertical Velocity after 3 seconds:
Initial vertical velocity, u = 120 m/s × sin(35.0°)
Acceleration due to gravity, a = -9.8 m/s² (negative sign because gravity acts downward)
Time, t = 3 s
Final vertical velocity, Vy = u + a×t
Resultant speed = √(horizontal velocity² + vertical velocity²)
Based on these calculations, the correct answer is D. 120 m/s, indicating that the projectile maintains its initial speed because the effect of gravity only changes the direction of the velocity, not its magnitude.
