A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 91 m. What is the speed of the ball just before it hits the catcher's glove 1.0 m above the ground? (g = 9.8 m/s2 and ignore air resistance)Select one:a.19 m/sb.38 m/sc.84 m/sd.42 m/s
Question
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 91 m. What is the speed of the ball just before it hits the catcher's glove 1.0 m above the ground? (g = 9.8 m/s2 and ignore air resistance)Select one:a.19 m/sb.38 m/sc.84 m/sd.42 m/s
Solution
To solve this problem, we can use the physics equation for the final velocity of an object in free fall, which is given by:
v_f = sqrt(v_i^2 + 2gh)
where:
- v_f is the final velocity (which we're trying to find),
- v_i is the initial velocity (which is 0 because the ball is dropped, not thrown),
- g is the acceleration due to gravity (9.8 m/s^2), and
- h is the height from which the ball is dropped (91 m - 1 m = 90 m, because we're asked for the speed just before the ball hits the catcher's glove, which is 1 m above the ground).
Substituting the given values into the equation, we get:
v_f = sqrt(0 + 29.890) v_f = sqrt(1764) v_f = 42 m/s
So, the speed of the ball just before it hits the catcher's glove is 42 m/s (option d).
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