A ball is thrown at an angle of 60 degrees with a speed of 15 m/s. What is the horizontal range of the ball?

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Q: A ball is thrown at an angle of 60 degrees with a speed of 15 m/s. What is the horizontal range of the ball?

Solution: Range (R) = (u^2 * sin(2θ)) / g = (15^2 * sin(120)) / 9.8 = 38.5 m.

Steps: 11

Step 1: Identify the initial speed (u) of the ball, which is 15 m/s.
Step 2: Identify the angle of projection (θ), which is 60 degrees.
Step 3: Convert the angle to radians if necessary, but here we can use degrees directly for the sine function.
Step 4: Calculate 2θ, which is 2 * 60 = 120 degrees.
Step 5: Calculate sin(120 degrees). This is equal to sin(180 - 60) = sin(60) = √3/2 ≈ 0.866.
Step 6: Use the formula for range (R): R = (u^2 * sin(2θ)) / g.
Step 7: Substitute the values into the formula: R = (15^2 * sin(120)) / 9.8.
Step 8: Calculate 15^2, which is 225.
Step 9: Multiply 225 by sin(120) (approximately 0.866): 225 * 0.866 ≈ 194.85.
Step 10: Divide 194.85 by g (approximately 9.8): 194.85 / 9.8 ≈ 19.87.
Step 11: The final result is the horizontal range of the ball, which is approximately 19.87 m.