Show all work please 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?
Accepted Solution
A:
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