A projectile is launched at an angle of 30 degrees with an initial velocity of 20 m/s. What is the maximum height reached by the projectile?

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Q: A projectile is launched at an angle of 30 degrees with an initial velocity of 20 m/s. What is the maximum height reached by the projectile?

Solution: Maximum height (H) = (u^2 * sin^2(θ)) / (2g) = (20^2 * (1/2)^2) / (2 * 9.8) = 10.2 m.

Steps: 12

Step 1: Identify the initial velocity (u) of the projectile, which is given as 20 m/s.
Step 2: Identify the angle of launch (θ), which is given as 30 degrees.
Step 3: Convert the angle from degrees to radians if necessary, but here we can use the sine function directly.
Step 4: Calculate the sine of the angle: sin(30 degrees) = 1/2.
Step 5: Use the formula for maximum height (H): H = (u^2 * sin^2(θ)) / (2g).
Step 6: Substitute the values into the formula: H = (20^2 * (1/2)^2) / (2 * 9.8).
Step 7: Calculate 20^2, which is 400.
Step 8: Calculate (1/2)^2, which is 1/4 or 0.25.
Step 9: Multiply 400 by 0.25 to get 100.
Step 10: Calculate 2 * 9.8, which is 19.6.
Step 11: Divide 100 by 19.6 to find the maximum height: 100 / 19.6 = 5.102 m.
Step 12: Round the answer to one decimal place, which gives us approximately 5.1 m.