An object is thrown vertically upwards with a speed of 30 m/s. How high will it rise before coming to rest momentarily?
Practice Questions
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Q1
An object is thrown vertically upwards with a speed of 30 m/s. How high will it rise before coming to rest momentarily?
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 upwards with a speed of 30 m/s. How high will it rise before coming to rest momentarily?
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: 9
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 acting downwards, so we use g = 9.8 m/s².
Step 4: Use the formula for height (h) which is h = (v² - u²) / (2g).
Step 5: Substitute the values into the formula: h = (0² - 30²) / (2 * -9.8).
Step 6: Calculate 30² which is 900, so now we have h = (0 - 900) / (2 * -9.8).
Step 7: Calculate 2 * -9.8 which is -19.6, so now we have h = -900 / -19.6.
Step 8: Divide -900 by -19.6 to find h. This gives h = 45.92 m.
Step 9: Round the answer to the nearest meter, which is approximately 45 m.
