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# PHYSICS PROBLEM HELP PLEASE?

Please show me how to do these i am so confused

A team of students competing in the Egg Drop competition at the Physics Olympics, drop their egg from a ledge, three floors above the ground.

1) If it takes the egg 2.4 seconds to reach the ground, calculate the height of the ledge in meters

2) How fast was the egg travelling the instant before it hit the ground

3) The student had planned to protect the egg by dropping it into a garbage can filled with confetti, 0.85m deep. Assuming that the egg stopped just as it reached the bottom of the can, calculate the rate of deceleration of the egg.

### 2 Answers

- electron1Lv 77 months agoFavorite Answer
1) If it takes the egg 2.4 seconds to reach the ground, calculate the height of the ledge in meters

Use the following equation to calculate the distance the egg falls in this time.

d = vi * t + ½ * a * t^2

Since the egg is dropped, its initial velocity is 0 m/s.

d = ½ * 9.8 * 2.4^2 = 28.224 meters

2) How fast was the egg travelling the instant before it hits the ground? Use the following equation to solve this problem.

vf = vi + a * t, vi = 0 m/s

vf = 9.8 * 2.4 = 23.52 m/s

3) The student had planned to protect the egg by dropping it into a garbage can filled with confetti, 0.85m deep. Assuming that the egg stopped just as it reached the bottom of the can, calculate the rate of deceleration of the egg.

Use the following equation to solve this problem.

vf^2 = vi^2 + 2 * a * d, vf = 0, vi = 23.52

0 = 23.52^2 + 2 * a * 0.85

a = -553.1904 ÷ 1.7

The acceleration is approximately -325 m/s^2. I hope this is helpful for you.

- VamanLv 77 months ago
height = 1/2 gt^2=28.22m

velocity when it touched the ground=9.8*2.4=23.52 m/s.

velocity becomes 0 at 0.85 m. Use v^2/2= a s. a is the deceleration rate and s is the distance. You get a= 23.52^2/2/0.85=325.4 m/s^2

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