Martin S. answered • 04/09/20
Patient, Relaxed PhD Molecular Biologist for Science and Math Tutoring
The object is at constant acceleration due to gravity, which is 32 feet per sec2 The acceleration is adding the same force at all times so velocity increases linearly, v = gt. The distance traveled is the time multiplied by the average of the initial velocity (zero) and the velocity at time = 3 seconds. So velocity at time 3 = 32 ft/sec2 x 3 sec = 96 ft/sec. The average velocity is one half that because the initial velocity is zero, so vavg = 48 ft/sec. Multiply that by the time . 3 seconds, and the distance traveled is 48 ft/sec x 3 sec = 144 feet. Only 6 feet to go!
Hope that helps
Martin S.
James, I believe you see the error in the stated problem. If air resistance is ignored, then we should only account for the object's motion as a consequence of gravity. Using 32 feet per sec2 for gravity, we can calculate that the object would travel 150 feet in just over 3 seconds. To calculate how far an object would travel after 5 seconds of freefall without air resistance, use 1/2 gt to find the average velocity after five seconds. That would be 1/2 x (5 x 32), or 90 feet per per second. Then multiply the average velocity by time ( 5 seconds) and we find that the object would travel 450 feet after five seconds of freefall. The object would fall 144 feet in the first three seconds, so in order for the problem to have been stated correctly, there would have to be a force partially counteracting gravity. Or perhaps the height of the tower was misstated.04/21/21
Martin S.
In order for the object to fall 150 feet in 5 seconds with only gravity acting upon it, then the average velocity would be 30 feet per second after 5 seconds. Back calculating from the equation v = 1/2 gt we can calculate g under those circumstances by rearranging the equation to get g = 2v/t, substituting 30 for v and 5 for t, then solving for g, we find that g would have to be 2 x 30 / 5 = 12 feet per second squared, which is 3/8 the gravity on Earth. Hmmm, the approximate gravity of Mars. Then using 12 ft/sec2 for g we can calculate that the average velocity would be 18 ft/sec after 3 seconds, and under those conditions the object would fall 54 feet in the first 3 seconds. If the problem had intended the student to consider gravity on another planet, then this is an interesting problem, indeed.04/21/21
James N.
so the object fall 150 feet, starting at rest, and hits the ground in 5 seconds. It seems, as stated, the object travels 144 feet in 3 seconds then suddenly slows to float down the remaining 6 feet in the remaining 2 seconds. oops04/21/21