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I have two math problems that I can't figure out.
1. What is the instantaneous velocity of a freely falling object 10 s after it is released from rest? What is the average velocity during this 10-s interval? How far will it fall during this time?

2. Kenny drops his physics book of a balcony. It hits the ground below 1.5 s later. With what speed does it hit? How high is the balcony? Ignore air drag.

Any help would be appreciated.

2007-08-13 03:10:39 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

1.
The final velocity v = u + at where u = 0 and a = g = 9.8m/s^2 and t = 10 s

v = 9.8 x 10 = 98m/s.

The average velocity = [u +v] / 2 = [0 + 98] / 2 = 49 m/s.

The distance traveled = average velocity x time
= 49 x 10
= 490 m

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2.

The final velocity v = u + at where u = 0 and a = g = 9.8m/s^2 and t = 1.5 s

v = 9.8 x 1.5 = 14.7 m/s.

The average velocity = [u +v] / 2 = [0 + 14.7] / 2 = 7.35 m/s.

The distance traveled = average velocity x time
= 7.35 x 1.5
= 11.025 m

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2007-08-13 07:35:42 · answer #1 · answered by Pearlsawme 7 · 0 0

start with v = g*t +v0 where g = accel. due to gravity = 9.8 m/s^2 and t = time and v0 = starting speed.

Then v = 9.8*t + 0 = 9.8 *(10s) = 98 m/s

Now average speed = distance/time

distance = x =1/2*g*t^2 +v0*t +x0 = 1/2*(9.8)*(10)^2 =49 m

average speed = 49 m/(10 s) = 4.9 m/s


2. Again v = g*t = 9.8 (1.5) m/s = 14.7 m/s

x = 1/2*g*t^2 = 1/2*9.8*(1.5)^2 = 11.03 m

2007-08-13 10:32:13 · answer #2 · answered by nyphdinmd 7 · 0 0

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