A projectile is launched with an initial velocity of 30 m/s at an angle of 45 degrees. What is the maximum height reached by the projectile? (2023)

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22.5 m
30 m
45 m
15 m

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Q: A projectile is launched with an initial velocity of 30 m/s at an angle of 45 degrees. What is the maximum height reached by the projectile? (2023)

Solution: Maximum height (H) = (u² * sin²θ) / (2g). Here, u = 30 m/s, θ = 45°, g = 9.81 m/s². H = (30² * (1/2)) / (2 * 9.81) = 22.5 m.

Steps: 12

Step 1: Identify the initial velocity (u) of the projectile, which is given as 30 m/s.
Step 2: Identify the launch angle (θ) of the projectile, which is given as 45 degrees.
Step 3: Convert the angle from degrees to radians if necessary, but for this calculation, we can use the sine of 45 degrees directly.
Step 4: Calculate the sine of the angle: sin(45°) = √2/2 = 1/√2 ≈ 0.7071. However, for the formula, we can use sin²(45°) = (1/√2)² = 1/2.
Step 5: Identify the acceleration due to gravity (g), which is approximately 9.81 m/s².
Step 6: Use the formula for maximum height (H) = (u² * sin²θ) / (2g).
Step 7: Substitute the values into the formula: H = (30² * (1/2)) / (2 * 9.81).
Step 8: Calculate 30², which is 900.
Step 9: Multiply 900 by (1/2), which gives 450.
Step 10: Calculate 2 * 9.81, which is 19.62.
Step 11: Divide 450 by 19.62 to find H: H = 450 / 19.62 ≈ 22.95 m.
Step 12: Round the answer to a reasonable number of significant figures, which gives us approximately 22.5 m.

A projectile is launched with an initial velocity of 30 m/s at an angle of 45 degrees. What is the maximum height reached by the projectile? (2023)

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