Show all work will give brainliest answer❕❗️❕❗️❕❗️❕❗️❕david drops a ball from a bridge at an initial height of 100 meters. (a) what is the height of the ball to the nearest tenth of a meter exactly 3 seconds after he releases the ball? (b) how many seconds after the ball is released will it hit the ground?

First, 9.8 (Gravity) times 3 (time) equals 29.4, which is the velocity after 3 seconds. The kinematic equation for change in position that uses the variables we have is:
delta x= (v)(t) -0.5(acceleration)(time)^2
delta x= 29.4 times (3) - 0.5 (9.8) times 9
delta x= 44.1
100 minus 44.1 equals 55.9, which is the answer for part a.

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PART B:
The kinematic equation for this is:
delta x= (initial velocity) times time plus 0.5 (a)(time)^2
100=(0)times(x) plus 0.5 (a)(time)^2
100=0.5(9.8)(x)^2
100=4.9x^2
100/4.9 is approxamitely 20.4.
The squareroot of this is approxamitely 4.5.
4.5 seconds

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