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QuestionA ball drops from a 53 ft building.
(a) Time to fall half the distance: t=t= sec\mathrm{sec}.
(b) Time to hit ground: t=t= sec\mathrm{sec}.

Studdy Solution

STEP 1

Assumptions1. The height of the building is53 ft. The ball is dropped from rest (initial velocity is0)
3. The acceleration due to gravity is -32. ft/sec² (negative because it's downward)
4. Air resistance is ignored5. The time it takes for the ball to fall half the distance and the full distance to the ground level is needed

STEP 2

We will use the equation of motion to solve this problem. The equation isd=vit+12at2d = v_i t + \frac{1}{2} a t^2where- dd is the distance the ball falls, - viv_i is the initial velocity (which is0 in this case), - aa is the acceleration (which is -32.2 ft/sec² due to gravity), and- tt is the time it takes for the ball to fall.

STEP 3

For part (a), we need to find the time it takes for the ball to fall half the distance to the ground level. The distance in this case is half the height of the building, which is53/2 =26.5 ft.
Substitute d=26.5d =26.5 ft, vi=0v_i =0 ft/sec, and a=−32.2a = -32.2 ft/sec² into the equation of motion26.5=0⋅t+12⋅−32.2⋅t226.5 =0 \cdot t + \frac{1}{2} \cdot -32.2 \cdot t^2

STEP 4

implify the equation to solve for tt26.=−16.1t226. = -16.1t^2

STEP 5

To isolate t2t^2, divide both sides of the equation by -16.1t2=26.5−16.1t^2 = \frac{26.5}{-16.1}

STEP 6

Calculate the value of t2t^2t2=−1.64658t^2 = -1.64658

STEP 7

Since time can't be negative, we take the square root of the absolute value of t2t^2 to find ttt=∣−1.64658∣t = \sqrt{|-1.64658|}

STEP 8

Calculate the value of ttt=1.283 sect =1.283 \, secSo, it will take approximately1.283 seconds for the ball to fall half the distance to the ground level.

STEP 9

For part (b), we need to find the time it takes for the ball to fall the full distance to the ground level. The distance in this case is the full height of the building, which is53 ft.
Substitute d=53d =53 ft, vi=v_i = ft/sec, and a=−32.2a = -32.2 ft/sec² into the equation of motion53=⋅t+2⋅−32.2⋅t253 = \cdot t + \frac{}{2} \cdot -32.2 \cdot t^2

STEP 10

implify the equation to solve for tt53=−16.t253 = -16.t^2

STEP 11

To isolate t^, divide both sides of the equation by -16.t=53−16.t^ = \frac{53}{-16.}

STEP 12

Calculate the value of t2t^2t2=−.29193t^2 = -.29193

STEP 13

Since time can't be negative, we take the square root of the absolute value of t2t^2 to find ttt=∣−3.29193∣t = \sqrt{|-3.29193|}

STEP 14

Calculate the value of ttt=.814 sect =.814 \, secSo, it will take approximately.814 seconds for the ball to fall to the ground level.

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