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David drops a ball from a bridge at an initial height of 70 meters what is the height of the ball to the nearest tenth of a meter exactly 2 seconds after he releases the ball?
Updated: 9/17/2023
Wiki User
∙ 14y agoBest Answer
Distance of fall in T seconds = 1/2 g T2
Distance of fall in 2 seconds = (1/2) (9.8) (2)2 = (4.9 x 4) = 19.6 meters
Height of this particular ball after 2 seconds = (70 - 19.6) = 50.4 meters
Wiki User
∙ 14y agoThis answer is:
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