I dont think and unassisted falling object would go that fast ?
Systematic experiments on freely falling objects and objects moving on inclined planes were carried out by Galileo Galilei (1564-1642). Freely falling objects are objects not supported by anything and not acted on by any forces except the gravitational force. Near the surface of the earth such objects are accelerating. This acceleration is due to the gravitational force acting between the objects and the earth. The direction of the gravitational acceleration vector is towards the center of the earth. Its magnitude decreases as one over the square of the distance from the center of the earth. The radius of the earth is 6368 km. If you climb a 1000 m high mountain, your distance from the center of the earth changes by (1/6368)´100 % = 0.016 % and the magnitude of the acceleration vector changes by (1/6368)2´100 % = 2.4´10-6 %. For all objects near the surface of the earth the distance from the center is nearly constant, and the magnitude of the gravitational acceleration vector is therefore approximately constant. We denote the gravitational acceleration vector by g. Its magnitude is g = 9.8 m/s2 and its direction is straight downward. Over small distances, when the curvature of the earth's surface can be neglected, the direction of the gravitational acceleration vector is also nearly constant.
Near the surface of the earth g is the same for all objects. All objects accelerate at approximately the same rate. Freely falling objects are therefore objects, which are moving with constant acceleration g.
Problems:
You drop a ball from a window on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you have a friend down on the street, who throws another ball upward with the same speed v. Your friend throws the ball upward at the same time you drop yours from the window. At some location the balls pass each other. Is this location at the halfway point between the window and the ground, above this point, or below this point? Solution
Both balls are accelerating. The ball thrown upward will reach the window at the same time the dropped ball reaches the ground. The speed of the dropped ball increases linearly with time. The ball moves with the slowest speed near the window and with the fastest speed near the point it hits the ground. In half the time it takes to reach the ground it covers less than half the distance and is still above the midpoint between window and ground. The speed of the ball thrown upwards decreases linearly with time. It moves fastest near the ground and slowest near the window. In half the time it takes to reach the window, it is already above the midway point. The two balls therefore meet above the midway point.
We can also work with our kinematic equations.
Assume the ball falls for 1 second. The speed of the falling ball as a function of time is v = 9.8 (m/s2) t and the distance traveled is d = (1/2) 9.8 (m/s2) t2. In one second the ball travels 4.9 m. The velocity of the falling ball as a function of time is v = -9.8 (m/s2) t j and its position as a function of time is r = (4.9 m - (1/2) 9.8 (m/s2) t2) j.
The velocity of the rising ball as a function of time is v = (9.8 (m/s) - 9.8 (m/s2) t) j and its position as a function of time is r = (9.8 (m/s) t - (1/2) 9.8( m/s2) t2) j. Graphs of velocity and position vs. time for the two balls are shown below.
Galileo experimented with balls rolling down inclined planes, in order to reduce the acceleration along the plane and thus reduce the rate of descent of the balls. Suppose the angle that the inclined plane makes with the horizontal is q. How would you expect the acceleration along the plane to decrease as q decreases. What specific trigonometric dependence on q would you expect for the acceleration? Solution:
Acceleration is a vector. The gravitational acceleration vector g is pointing straight down. Vectors can be written in terms of their components along the axes of a coordinate system. If we choose our coordinate system as shown, then the x-component of g is parallel and the y-component of g is perpendicular to the surface of the inclined plane.
We have gx = gsinq. The object is constrained to move on the surface of the plane, and therefore will accelerate in the x-direction. We expect the magnitude of the acceleration to be equal to gx and to depend on q as sinq .
A pebble is dropped into a water well, and the splash is heard 16 s later. What is the approximate distance from the rim of the well to the water's surface? Solution:
If we assume that the pebble is a freely falling object, then the distance covered by the pebble in time t is d = (1/2)gt2, where g = 9.8 m/s2 is the gravitational acceleration. If the pebble were falling for 16 s, the d = (1/2)9.8 (m/s2) 256 s2 = 1254 m and its speed at that time would be v = gt = 9.8 (m/s2) 16 s = 156 m/s = 351 mph. However, since the speed of sound is 343 m/s, we would only hear the splash after 19.6 s. The pebble must therefore be falling for a shorter time and travel a shorter distance. It must fall for somewhere between 13s and 14 s. Let us assume it falls for 13.5 s. Then d = 893 m. Sound covers this distance in 2.6 s. So 890 m is roughly right, as long as we are justified to neglect air resistance.
A ball is thrown directly downward with an initial speed of 8 m/s from a height of 30m. After what time interval does the ball strike the ground? Solution:
We have motion in one dimension with constant acceleration. Put the origin of your coordinate system at the position of the ball at t = 0 and let the x-axis point straight downward. Then we know that
xi = 0, vxi = 8 m/s, ti = 0,
xf = 30 m, vxf = ?, tf = ?,
ax = 9.8 m/s2.
We are asked to solve for t = tf.
We have xf - xi = vxiDt + (1/2)axDt2, 30 m = 8 (m/s) t + 4.9 (m/s2) t2. We rewrite this equation as t2 + (8 s/4.9) t - (30 s2/4.9) = 0, or t2 + (1.63 s)t - 6.12 s2 = 0. This is a quadratic equation with two solutions.
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Only the + sign makes sense. We find t = 1.78s
A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in contact with the pavement, the lower side of the ball is temporarily flattened. Assume that the maximum depth of the dent in the ball is about 1cm. Make an order of magnitude estimate of the maximum acceleration of the ball. State your assumptions, the quantities you estimate, and the values you estimate for them. Solution:
The ball falls from chest height, so it falls for a distance of approximately 1.3m under constant acceleration g. It falls for a time , and contacts the ground with speed v = gt = 5 m/s.
The center of the ball now moves another centimeter before the ball stops and v = 0. We can use vxf2 - vxi2 = 2ax(xf - xi) to find the average acceleration of the ball during this time interval.
-(5 m/s)2 = 2 ax 0.01 m. ax = -1205 m/s2.
A ball rolls up an incline, and then rolls back down to its initial position. Which of the following graphs best represents the velocity of the ball as a function of time?
Solution
If we orient the x-axis of our coordinate system so it points uphill along the incline, then at t = 0 vx is positive. The velocity vx decreases linearly since ax points into the negative x-direction. When the ball reaches its highest point, vx = 0, when it rolls back down the incline vx is negative. The graph (C) best represents the velocity of the ball as a function of time.
ugh what! he would only be able to reach terminal velocity which (I think is what he is saying above), would not be able to hit the speed of sound! just my$0.01
