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QuestionA ball is dropped from a 94 ft building. Find the time to fall half the distance and to ground level: (a) t=t= sec, (b) t=t= sec.

Studdy Solution

STEP 1

Assumptions1. The height of the building is94 ft. . The ball is dropped from the top of the building.
3. The acceleration due to gravity is -32. ft/sec². (This is negative because it acts in the downward direction.)
4. The initial velocity of the ball is0 ft/sec, as it is simply dropped, not thrown.
5. The time it takes for the ball to fall half the distance and to the ground level is what we need to find.

STEP 2

We can use the equation of motion to find the time it takes for the ball to fall a certain distance. The equation of motion is given byd=vit+12at2d = v_i t + \frac{1}{2} a t^2where- dd is the distance, - viv_i is the initial velocity, - aa is the acceleration, and- tt is the time.

STEP 3

First, let's find the time it takes for the ball to fall half the distance to the ground level. Half the height of the building is 94/2=4794/2 =47 ft. So, we need to solve the equation of motion for d=47d =47 ft.
47=0⋅t+12⋅−32.2⋅t247 =0 \cdot t + \frac{1}{2} \cdot -32.2 \cdot t^2

STEP 4

implify the equation to find the value of t2t^2.
t2=2⋅47−32.2t^2 = \frac{2 \cdot47}{-32.2}

STEP 5

Calculate the value of t2t^2.
t2=2⋅47−32.2=−2.919t^2 = \frac{2 \cdot47}{-32.2} = -2.919

STEP 6

Since time cannot be negative, we take the absolute value of t2t^2 and then find the square root to get tt.
t=∣−2.919∣t = \sqrt{|-2.919|}

STEP 7

Calculate the value of tt.
t=∣−2.919∣=1.708 sect = \sqrt{|-2.919|} =1.708 \, secSo, it will take approximately1.708 seconds for the ball to fall half the distance to the ground level.

STEP 8

Next, let's find the time it takes for the ball to fall to the ground level. We need to solve the equation of motion for d=94d =94 ft.
94=0⋅t+12⋅−32.2⋅t294 =0 \cdot t + \frac{1}{2} \cdot -32.2 \cdot t^2

STEP 9

implify the equation to find the value of t2t^2.
t2=2⋅94−32.2t^2 = \frac{2 \cdot94}{-32.2}

STEP 10

Calculate the value of t2t^2.
t2=2⋅94−32.2=−5.838t^2 = \frac{2 \cdot94}{-32.2} = -5.838

STEP 11

Since time cannot be negative, we take the absolute value of t^ and then find the square root to get tt.
t=∣−5.838∣t = \sqrt{|-5.838|}

STEP 12

Calculate the value of tt.
t=∣−5.838∣=2.416 sect = \sqrt{|-5.838|} =2.416 \, secSo, it will take approximately2.416 seconds for the ball to fall to the ground level.

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