Physics · Motion in Two Dimensions
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If a projectile is launched with 3m/s velocity at 60-degree angle then at the highest point its horizontal velocity is:
- A
3 m/s
- B
2m/s
- C
1.5 m/s
- D
1.8 m/s
At the highest point of the projectile's trajectory, its horizontal velocity will be equal to the initial horizontal velocity. Therefore, the horizontal velocity of the projectile at the highest point will be:
v*cosθ = 3*cos(60) = 1.5 m/s
where v is the initial velocity and theta is the launch angle.
At the highest point of the projectile's trajectory, its horizontal velocity will be equal to the initial horizontal velocity. Therefore, the horizontal velocity of the projectile at the highest point will be:
v*cosθ = 3*cos(60) = 1.5 m/s
where v is the initial velocity and theta is the launch angle.
At the highest point of the projectile's trajectory, its horizontal velocity will be equal to the initial horizontal velocity. Therefore, the horizontal velocity of the projectile at the highest point will be:
v*cosθ = 3*cos(60) = 1.5 m/s
where v is the initial velocity and theta is the launch angle.
At the highest point of the projectile's trajectory, its horizontal velocity will be equal to the initial horizontal velocity. Therefore, the horizontal velocity of the projectile at the highest point will be:
v*cosθ = 3*cos(60) = 1.5 m/s
where v is the initial velocity and theta is the launch angle.
At the highest point of the projectile's trajectory, its horizontal velocity will be equal to the initial horizontal velocity. Therefore, the horizontal velocity of the projectile at the highest point will be:
v*cosθ = 3*cos(60) = 1.5 m/s
where v is the initial velocity and theta is the launch angle.
Tagged under Physics · Motion in Two Dimensions · 2021