**Below are the best information and knowledge on the subject find the velocity of the coin when it hits the ground compiled and compiled by our own team thienmaonline:**

## 1. SOLUTION: a coin is dropped from a height of 750 feet. the height,s,at time,t, is given by s=-16t^2+750. a. find the average velocity on the interval[1,3] b. find the instantaneous velo

**Author: ** www.youtube.com

**Date Submitted: ** 03/08/2020 12:18 AM

**Average star voting: ** 3 ⭐ ( 66811 reviews)

**Summary: **

**Match with the search results: ** www.youtube.com › watch…. read more

## 2. a silver dollar is dropped from the top of a building | Wyzant Ask An Expert

**Author: ** socratic.org

**Date Submitted: ** 07/10/2021 09:21 AM

**Average star voting: ** 3 ⭐ ( 92687 reviews)

**Summary: **

**Match with the search results: ** 6.85 sec. Explanation: At ground level, s=0. s=0⇒−16t2+750=0 ∴16t2=750 ∴t2=75016 ∴t2=3758 ∴t=±√3758 ∴t=±6.85 (2dp)….. read more

## 3. A dollar coin is thrown from the top of a tall building. Its height (in feet) after t seconds is given by s(t) = 1024 – 16t^2 (a) Find the velocity function v(t) for the coin. (b) Find when the coin h

**Author: ** www.quora.com

**Date Submitted: ** 10/05/2019 05:55 PM

**Average star voting: ** 4 ⭐ ( 65960 reviews)

**Summary: ** Answer to: A dollar coin is thrown from the top of a tall building. Its height (in feet) after t seconds is given by s(t) = 1024 – 16t^2 (a) Find the…

**Match with the search results: ** The final velocity of the coin is solved by multiplying the elapsed time t by the acceleration due to gravity g. V_final = g * t = 9.81 m/s^2 * 2.48 s = 24.3 m/ ……. read more

## 4. Top 3 find the velocity of the coin when it gets to 50 ft in 2022 – Gấu Đây

**Author: ** www.webassign.net

**Date Submitted: ** 01/08/2020 05:32 PM

**Average star voting: ** 4 ⭐ ( 68818 reviews)

**Summary: **

**Match with the search results: ** The value v = +2.15 m/s is the velocity of the coin on the upward trip, and v = –2.15 m/s is the velocity on the downward trip. The speed in both cases is ……. read more

## 5. SOLVED:A coin is dropped from a hot-air balloon that is 300 m above the ground and rising at 10.0 m/s upward. For the coin, find (a) the maximum height reached, (b) its position and velocity 4.00 s af

**Author: ** thaiphuongthuy.com

**Date Submitted: ** 09/28/2020 03:38 AM

**Average star voting: ** 5 ⭐ ( 18160 reviews)

**Summary: ** Okay, so for this coin dropped on about hot air balloon, the hot air balloon itself is moving upward at 10 meters per second and the coin has dropped when it’s 300 meters above the ground. Now the movement of the hot air balloon is the same movement as the coin and tell it’s released. And so that means This coin also has an initial velocity of 10 meters per second upward once it’s released. No, the first, um, question asked, What is it? Maximum height that it gets you. So if it’s traveling upward at 10 meters per second, even once it’s let go, it’s going to continue traveling upward as gravity slows it down until it momentarily stops and then plummets downward. So we used the equation. B squared equals initial velocity squared plus two times eight times X, and the final velocity in this case is zero because of its the coins maximum height. That’s where it momentarily comes to a stop before it starts to fall. So when we solve that, we find that it changes its position by 5.1 meters. Now that means compared to the ground if it changed its position 5.1 meter starting at 300 meters. It’s total maximum height is 305 meters off the ground. The second part of this problem says at four seconds what is its position in the air and how fast is it going? So now we have to look at its initial velocity is still positive 10 meters per second and it still starts 300 meters above the ground. But now we have a time for four sites. So to solve for the position we use, position equals initial position plus initial velocity, times time plus 1/2 acceleration times, Times Square. And we have all of these pieces. So we plug it in. We find that after four seconds, this coin is 262 meters above the ground to solve for velocity. That’s pretty straightforward. We just use the equation. Velocity equals initial velocity plus acceleration, times, time and when we put those numbers and we find that it has a negative velocity of 29.2 meters circuit for the final problem, it wants to know how long is this coin in the air total before it hits the ground. So again we have all of our information the initial, um, velocity and initial position acceleration. In this time, the final position is zero meters because it wants to know how long it takes to get to the ground. So we’re gonna use the same equation we did in the previous problem. Position equals initial position, plus initial velocity, times time plus 1/2 acceleration times, time squared. The problem is, this time we’re looking for time and time is our independent variable in this quadratic equation. So to solve for tea, you have to use the quadratic formula to solve. This is going to give you two different possible answers, one of which is positive. And one is negative because we’re solving for time. A negative answer doesn’t make any sense. So the correct answer here is time is 8.91 seconds right to him.

**Match with the search results: ** 5. A dollar coin is thrown from the top of a tall building. Its height (in feet) after t seconds is given by s(t) = 1024 – 16t^2 (a) Find the velocity function ……. read more

## 6. SOLVED:a coin is dropped from a bridge that is 6 ft (1 ft = 0.3 meters) if accleration is 9.8m/s/s, how long does it take for the coin to hit the ground. once you find time, calculate its velocity bef

**Author: ** www.algebra.com

**Date Submitted: ** 10/08/2019 02:25 PM

**Average star voting: ** 3 ⭐ ( 45013 reviews)

**Summary: ** Hello guys. And this problem we have the height of drop for the queen as six ft. Okay, so to solve this problem first we have to convert fit in tomatoes. So we have to multiply with the unit conversion factor. That is 0.3 m poor feet. Therefore the height of all equals 1.8 m. No, for the party under gravity, for the motion starting from rest, the height of fall equals a half times the gravitational acceleration into the square of time of fall. So the time of fall equals square root off. Two times they hide a fall over the gravitational acceleration. Now substitute the values to find out the time of fall that equals two times 1.8 m over 9.8 m/s square are the time of fall. For the coin equals zero point six 0.606 second are And one significant figure. The time of four list 0.6 seconds. No party. The funny velocity equals the gravitational acceleration into the time of fall. Are that equals 9.8 m per second squared times zero six seconds Are the final velocity of the coin when it hit the ground. This 5.9 needles or second along the downward direction

**Match with the search results: ** a. find the average velocity on the interval[1,3] b. find the instantaneous velocity when t=3 c. find the velocity of the coin when it hits the ground…. read more

## 7. Falling Objects – College Physics

**Author: ** www.wyzant.com

**Date Submitted: ** 07/20/2019 06:02 AM

**Average star voting: ** 4 ⭐ ( 42305 reviews)

**Summary: **

**Match with the search results: ** (d) Find the time required for the coin to reach the ground level. … (e) Find the velocity of the coin at impact….. read more

## 8. Calculus and Physics with Velocity

**Author: ** study.com

**Date Submitted: ** 11/20/2020 08:45 PM

**Average star voting: ** 4 ⭐ ( 34849 reviews)

**Summary: ** A coin is dropped from a height of 750 feet. The height, s, (measured in feet), at time, t (measured in seconds), is given by s= -16t2 + 750.

a) Find the…

**Match with the search results: ** (b) Find when the coin hits the ground. (c) What is the coin’s velocity at impact? (Remember to include the units!) Free-fall Equation for ……. read more

## 9.

**Author: ** brainly.com

**Date Submitted: ** 01/23/2020 07:08 AM

**Average star voting: ** 5 ⭐ ( 30044 reviews)

**Summary: **

**Match with the search results: ** The speed of the coin as it hits the ground is 14.715 m/s. We can solve the problem above using the equation of acceleration under gravity….. read more