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Difficulty: EasyProjectile Motion

A cannonball is fired from level ground with an initial speed of 20 m/s20\text{ m/s} at an angle of 30∘30^\circ above the horizontal. Taking the acceleration due to gravity g=10 m/s2g = 10\text{ m/s}^2, what is the total time of flight of the cannonball before it returns to ground level?

  1. 2.0 s2.0\text{ s}Answer
  2. B
    1.0 s1.0\text{ s}
  3. C
    4.0 s4.0\text{ s}
  4. D
    0.5 s0.5\text{ s}

Answer

2.0 s2.0\text{ s}
The correct answer of 2.0 s2.0\text{ s} is determined by resolving the initial speed into its vertical component uy=20sin⁡30∘=10 m/su_y = 20 \sin 30^\circ = 10\text{ m/s} and using the total time of flight formula T=2usin⁡θg=2(10)10=2.0 sT = \frac{2 u \sin \theta}{g} = \frac{2(10)}{10} = 2.0\text{ s}.

Step-by-Step Solution

1
Calculate the vertical component of the initial velocity (uyu_y).
uy=usin⁡θ=20×sin⁡30∘=20×0.5=10 m/su_y = u \sin \theta = 20 \times \sin 30^\circ = 20 \times 0.5 = 10\text{ m/s}
Only the vertical component of velocity determines the time the projectile remains in the air.
2
Calculate the total time of flight (TT) using the kinematic formula for full trajectory.
T=2uyg=2×1010=2.0 sT = \frac{2 u_y}{g} = \frac{2 \times 10}{10} = 2.0\text{ s}
The total time of flight includes both the time to ascend to peak height and descend back to the launch height under gravity gg.

Key Concept

Total Time of Flight in Projectile Motion
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Topics

Projectile Motion

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