A ball is thrown vertically upward with an initial velocity of 20 m/s. How high will it rise before coming to a momentary stop? (g = 10 m/s²)

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Q: A ball is thrown vertically upward with an initial velocity of 20 m/s. How high will it rise before coming to a momentary stop? (g = 10 m/s²)

Solution: Using the formula v² = u² + 2as, where v = 0, u = 20 m/s, and a = -g = -10 m/s², we get 0 = (20)² + 2(-10)s, solving gives s = 20 m.

Steps: 8

Step 1: Identify the initial velocity (u) of the ball, which is given as 20 m/s.
Step 2: Identify the final velocity (v) of the ball at the highest point, which is 0 m/s (the ball stops momentarily).
Step 3: Identify the acceleration (a) due to gravity, which is -10 m/s² (negative because it acts downward).
Step 4: Use the formula v² = u² + 2as. Substitute the values: 0 = (20)² + 2(-10)s.
Step 5: Simplify the equation: 0 = 400 - 20s.
Step 6: Rearrange the equation to solve for s: 20s = 400.
Step 7: Divide both sides by 20 to find s: s = 400 / 20.
Step 8: Calculate s: s = 20 m.