A projectile is launched with an initial velocity of 30 m/s at an angle of 30° to the horizontal. What is the horizontal range of the projectile? (Take g = 10 m/s²)
Practice Questions
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Q1
A projectile is launched with an initial velocity of 30 m/s at an angle of 30° to the horizontal. What is the horizontal range of the projectile? (Take g = 10 m/s²)
90 m
75 m
100 m
120 m
Range R = (u² * sin(2θ))/g = (30² * sin(60°))/10 = (900 * √3/2)/10 = 45√3 m ≈ 90 m.
Questions & Step-by-step Solutions
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Q
Q: A projectile is launched with an initial velocity of 30 m/s at an angle of 30° to the horizontal. What is the horizontal range of the projectile? (Take g = 10 m/s²)
Solution: Range R = (u² * sin(2θ))/g = (30² * sin(60°))/10 = (900 * √3/2)/10 = 45√3 m ≈ 90 m.
Steps: 9
Step 1: Identify the initial velocity (u) of the projectile, which is given as 30 m/s.
Step 2: Identify the launch angle (θ), which is given as 30 degrees.
Step 3: Calculate the value of 2θ, which is 2 * 30° = 60°.
Step 4: Use the sine function to find sin(60°). The value of sin(60°) is √3/2.
Step 5: Substitute the values into the range formula: R = (u² * sin(2θ))/g.
Step 6: Calculate u², which is 30² = 900.
Step 7: Substitute u² and sin(60°) into the formula: R = (900 * (√3/2))/10.
Step 8: Simplify the equation: R = (900 * √3/2) / 10 = 90√3.
Step 9: Calculate the approximate value of R: 90√3 is approximately 90 m.
