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sveticcg [70]
3 years ago
8

A rock is thrown straight down at 7.2 m/s off a bridge that is 45 m above the ground. Find the

Physics
1 answer:
iragen [17]3 years ago
5 0

The rock's height <em>y</em> at time <em>t</em> is given by

<em>y</em> = 45 m + (7.2 m/s) <em>t</em> - 1/2 <em>g</em> <em>t </em>²

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Set <em>y</em> = 0 and solve for <em>t</em> :

0 = 45 m + (7.2 m/s) <em>t</em> - 1/2 <em>g</em> <em>t </em>²   →   <em>t</em> ≈ 3.9 s

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When driver see the child standing on road his speed is 20 m/s

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3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

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What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

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                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

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- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

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