An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to a momentary stop?
Practice Questions
1 question
Q1
An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to a momentary stop?
45 m
60 m
90 m
135 m
Using the formula: h = (v² - u²) / (2g), where v = 0 m/s (at the highest point), u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.
Questions & Step-by-step Solutions
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Q
Q: An object is thrown vertically upward with a speed of 30 m/s. How high will it rise before coming to a momentary stop?
Solution: Using the formula: h = (v² - u²) / (2g), where v = 0 m/s (at the highest point), u = 30 m/s, g = 9.8 m/s². h = (0 - 30²) / (2 * -9.8) = 45.92 m, approximately 45 m.
Steps: 8
Step 1: Identify the initial speed (u) of the object. In this case, u = 30 m/s.
Step 2: Recognize that at the highest point, the final speed (v) will be 0 m/s.
Step 3: Understand that the acceleration due to gravity (g) is approximately 9.8 m/s². Since the object is moving upward, we will use -g in our calculations.
Step 4: Use the formula for height (h) which is h = (v² - u²) / (2g).
Step 5: Substitute the values into the formula: h = (0 - 30²) / (2 * -9.8).
Step 6: Calculate 30², which is 900. So now we have h = (-900) / (-19.6).
Step 7: Divide -900 by -19.6 to find h. This gives h = 45.92 m.
Step 8: Round the answer to the nearest whole number, which is approximately 45 m.
