QuestionA ball drops from a 94 ft building. (Round to 3 decimal places.) (a) Time to fall half distance? $t=1.8$ (b) Time to reach ground? $t=2.4$

Studdy Solution

STEP 1

Assumptions1. The height of the building is94 ft. . The acceleration due to gravity is approximately32. ft/s².
3. The ball is dropped from rest (initial velocity is0).
4. Air resistance and other factors are ignored.

STEP 2

We will use the equation of motion to solve this problem. The equation is $h = h0 + v0t - \frac{1}{2}gt^2$ where- $h$ is the final height, - $h0$ is the initial height, - $v0$ is the initial velocity, - $g$ is the acceleration due to gravity, and- $t$ is the time.

STEP 3

To find the time it takes for the ball to fall half the distance, we set $h$ to half the height of the building. So, $h =94/2 =47$ ft.

STEP 4

Now, we substitute the known values into the equation of motion $47 =94 +0 \cdot t - \frac{1}{2} \cdot32.2 \cdot t^2$

STEP 5

olving the equation for $t^2$ gives $t^2 = \frac{94 -47}{0.5 \cdot32.2}$

STEP 6

Now, we calculate the value of $t^2$ $t^2 = \frac{47}{16.1}$

STEP 7

To find $t$ , we take the square root of $t^2$ $t = \sqrt{\frac{47}{16.1}}$

STEP 8

Calculating the value of $t$ gives $t \approx1.72 \, \text{sec}$ So, it will take approximately1.72 seconds for the ball to fall half the distance to the ground.

STEP 9

To find the time it takes for the ball to fall to the ground, we set $h$ to (because the ball reaches the ground). So, the equation of motion becomes $=94 + \cdot t - \frac{}{2} \cdot32.2 \cdot t^2$

STEP 10

olving this equation for $t^2$ gives $t^2 = \frac{94}{0.5 \cdot32.2}$

STEP 11

Now, we calculate the value of t^ $t^ = \frac{94}{16.}$

STEP 12

To find $t$ , we take the square root of $t^2$ $t = \sqrt{\frac{94}{16.}}$

STEP 13

Calculating the value of $t$ gives $t \approx2.44 \, \text{sec}$ So, it will take approximately2.44 seconds for the ball to fall to the ground.

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QuestionA ball drops from a 94 ft building. (Round to 3 decimal places.) (a) Time to fall half distance? $t=1.8$ (b) Time to reach ground? $t=2.4$

Studdy Solution

STEP 1

Assumptions1. The height of the building is94 ft. . The acceleration due to gravity is approximately32. ft/s².
3. The ball is dropped from rest (initial velocity is0).
4. Air resistance and other factors are ignored.

STEP 2

We will use the equation of motion to solve this problem. The equation is $h = h0 + v0t - \frac{1}{2}gt^2$ where- $h$ is the final height, - $h0$ is the initial height, - $v0$ is the initial velocity, - $g$ is the acceleration due to gravity, and- $t$ is the time.

STEP 3

To find the time it takes for the ball to fall half the distance, we set $h$ to half the height of the building. So, $h =94/2 =47$ ft.

STEP 4

Now, we substitute the known values into the equation of motion $47 =94 +0 \cdot t - \frac{1}{2} \cdot32.2 \cdot t^2$

STEP 5

olving the equation for $t^2$ gives $t^2 = \frac{94 -47}{0.5 \cdot32.2}$

STEP 6

Now, we calculate the value of $t^2$ $t^2 = \frac{47}{16.1}$

STEP 7

To find $t$ , we take the square root of $t^2$ $t = \sqrt{\frac{47}{16.1}}$

STEP 8

Calculating the value of $t$ gives $t \approx1.72 \, \text{sec}$ So, it will take approximately1.72 seconds for the ball to fall half the distance to the ground.

STEP 9

To find the time it takes for the ball to fall to the ground, we set $h$ to (because the ball reaches the ground). So, the equation of motion becomes $=94 + \cdot t - \frac{}{2} \cdot32.2 \cdot t^2$

STEP 10

olving this equation for $t^2$ gives $t^2 = \frac{94}{0.5 \cdot32.2}$

STEP 11

Now, we calculate the value of t^ $t^ = \frac{94}{16.}$

STEP 12

To find $t$ , we take the square root of $t^2$ $t = \sqrt{\frac{94}{16.}}$

STEP 13

Calculating the value of $t$ gives $t \approx2.44 \, \text{sec}$ So, it will take approximately2.44 seconds for the ball to fall to the ground.

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QuestionA ball drops from a 94 ft building. (Round to 3 decimal places.) (a) Time to fall half distance? t=1.8 t=1.8 t=1.8 (b) Time to reach ground? t=2.4 t=2.4 t=2.4

Studdy Solution

STEP 1

Assumptions1. The height of the building is94 ft. . The acceleration due to gravity is approximately32. ft/s².3. The ball is dropped from rest (initial velocity is0).4. Air resistance and other factors are ignored.

STEP 2

We will use the equation of motion to solve this problem. The equation ish=h0+v0t−12gt2h = h0 + v0t - \frac{1}{2}gt^2h=h0+v0t−21​gt2where- hhh is the final height, - h0h0h0 is the initial height, - v0v0v0 is the initial velocity, - ggg is the acceleration due to gravity, and- ttt is the time.

STEP 3

To find the time it takes for the ball to fall half the distance, we set hhh to half the height of the building. So, h=94/2=47h =94/2 =47h=94/2=47 ft.

STEP 4

Now, we substitute the known values into the equation of motion47=94+0⋅t−12⋅32.2⋅t247 =94 +0 \cdot t - \frac{1}{2} \cdot32.2 \cdot t^247=94+0⋅t−21​⋅32.2⋅t2

STEP 5

olving the equation for t2t^2t2 givest2=94−470.5⋅32.2t^2 = \frac{94 -47}{0.5 \cdot32.2}t2=0.5⋅32.294−47​

STEP 6

Now, we calculate the value of t2t^2t2t2=4716.1t^2 = \frac{47}{16.1}t2=16.147​

STEP 7

To find ttt, we take the square root of t2t^2t2t=4716.1t = \sqrt{\frac{47}{16.1}}t=16.147​​

STEP 8

Calculating the value of ttt givest≈1.72 sect \approx1.72 \, \text{sec}t≈1.72secSo, it will take approximately1.72 seconds for the ball to fall half the distance to the ground.

STEP 9

To find the time it takes for the ball to fall to the ground, we set hhh to (because the ball reaches the ground). So, the equation of motion becomes=94+⋅t−2⋅32.2⋅t2 =94 + \cdot t - \frac{}{2} \cdot32.2 \cdot t^2=94+⋅t−2​⋅32.2⋅t2

STEP 10

olving this equation for t2t^2t2 givest2=940.5⋅32.2t^2 = \frac{94}{0.5 \cdot32.2}t2=0.5⋅32.294​

STEP 11

Now, we calculate the value of t^t=9416.t^ = \frac{94}{16.}t=16.94​

STEP 12

To find ttt, we take the square root of t2t^2t2t=9416.t = \sqrt{\frac{94}{16.}}t=16.94​​

STEP 13

Calculating the value of ttt givest≈2.44 sect \approx2.44 \, \text{sec}t≈2.44secSo, it will take approximately2.44 seconds for the ball to fall to the ground.

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QuestionA ball drops from a 94 ft building. (Round to 3 decimal places.) (a) Time to fall half distance? $t=1.8$ (b) Time to reach ground? $t=2.4$

Assumptions1. The height of the building is94 ft. . The acceleration due to gravity is approximately32. ft/s².
3. The ball is dropped from rest (initial velocity is0).
4. Air resistance and other factors are ignored.

We will use the equation of motion to solve this problem. The equation is $h = h0 + v0t - \frac{1}{2}gt^2$ where- $h$ is the final height, - $h0$ is the initial height, - $v0$ is the initial velocity, - $g$ is the acceleration due to gravity, and- $t$ is the time.

To find the time it takes for the ball to fall half the distance, we set $h$ to half the height of the building. So, $h =94/2 =47$ ft.

Now, we substitute the known values into the equation of motion $47 =94 +0 \cdot t - \frac{1}{2} \cdot32.2 \cdot t^2$

olving the equation for $t^2$ gives $t^2 = \frac{94 -47}{0.5 \cdot32.2}$

Now, we calculate the value of $t^2$ $t^2 = \frac{47}{16.1}$

To find $t$ , we take the square root of $t^2$ $t = \sqrt{\frac{47}{16.1}}$

Calculating the value of $t$ gives $t \approx1.72 \, \text{sec}$ So, it will take approximately1.72 seconds for the ball to fall half the distance to the ground.

To find the time it takes for the ball to fall to the ground, we set $h$ to (because the ball reaches the ground). So, the equation of motion becomes $=94 + \cdot t - \frac{}{2} \cdot32.2 \cdot t^2$

olving this equation for $t^2$ gives $t^2 = \frac{94}{0.5 \cdot32.2}$

Now, we calculate the value of t^ $t^ = \frac{94}{16.}$

To find $t$ , we take the square root of $t^2$ $t = \sqrt{\frac{94}{16.}}$

Calculating the value of $t$ gives $t \approx2.44 \, \text{sec}$ So, it will take approximately2.44 seconds for the ball to fall to the ground.