A projectile is launched at an angle of 30° with the horizontal with an initial velocity of 40 m/s. What is the horizontal range of the projectile? (Take g = 10 m/s²)
Practice Questions
1 question
Q1
A projectile is launched at an angle of 30° with the horizontal with an initial velocity of 40 m/s. What is the horizontal range of the projectile? (Take g = 10 m/s²)
160 m
200 m
80 m
120 m
Range R = (u² * sin(2θ))/g = (40² * sin(60°))/10 = (1600 * √3/2)/10 = 80√3 m ≈ 138.56 m.
Questions & Step-by-step Solutions
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Q
Q: A projectile is launched at an angle of 30° with the horizontal with an initial velocity of 40 m/s. What is the horizontal range of the projectile? (Take g = 10 m/s²)
Solution: Range R = (u² * sin(2θ))/g = (40² * sin(60°))/10 = (1600 * √3/2)/10 = 80√3 m ≈ 138.56 m.
Steps: 9
Step 1: Identify the given values. The initial velocity (u) is 40 m/s, the angle (θ) is 30°, and the acceleration due to gravity (g) is 10 m/s².
Step 2: Use the formula for the range of a projectile: R = (u² * sin(2θ)) / g.
Step 3: Calculate sin(2θ). Since θ is 30°, 2θ is 60°. Therefore, sin(60°) = √3/2.
Step 4: Substitute the values into the range formula: R = (40² * sin(60°)) / 10.
Step 5: Calculate 40², which is 1600.
Step 6: Substitute sin(60°) into the equation: R = (1600 * √3/2) / 10.
Step 7: Simplify the equation: R = (1600 * √3) / 20.
Step 8: Calculate 1600 / 20, which equals 80. So, R = 80√3.
Step 9: To find the approximate value, calculate 80 * √3, which is approximately 138.56 m.
