A boy throws a rock horizontally from an elevated position 16 feet above the ground (perhaps from a cliff's edge). The rock travels a horizontal distance of 40 feet. How fast did the boy throw the rock? (Again, consider time.)
a) 16 ft/s
b) 40 ft/s
c) 43 ft/s
d) 56 ft/s
e) Cannot be solved with the information given
2007-10-29 16:32:21 · 3 answers · asked by ? 6 in Science & Mathematics ➔ Physics
The answer is: b. We know that speed = distance traveled/time taken and we are given that the rock travels 40 feet. Since the rock was thrown exactly horizontally with no initial component of velocity in the up or down direction, the time taken is the same time that would be required for the rock to fall 16 feet if it were simply dropped from rest. This time is one second, so the stone's horizontal velocity must be 40 ft/s.
2007-10-31 15:55:00 · update #1
B's the answer.
A rock dropped from a height of 16 ft, assuming acceleration of gravity at 32 ft*sec^-2 and no air resistance, will reach the earth's surface in one second. Therefore, since the horizontal and vertical velocities are independent of each other (at least at the speeds in question), a rock thrown horizontally will fall vertically at the same rate as a dropped rock, i.e., 16 ft in one second.
We know that the thrown rock traveled horizontally 40 ft., and that it must have done so during the one second interval before it hit the ground. Presume there were no outside forces acting on the rock in the horizontal direction (an object in motion will tend to stay in motion), the horizontal velocity must have been 40 ft/sec.
Let me restate my presumptions in solving this problem: (1) That time and all horizontal and vertical distances are all independent of each other, (2) The only external force acting on the system is gravity, acting vertically, at 32 ft*sec^-2, and (3) When the rock hits the ground, it comes to an immediate stop. ;-)
2007-10-29 18:07:27 · answer #1 · answered by Frst Grade Rocks! Ω 7 · 1⤊ 0⤋
first, we need to find time
-use the formula t = sqrt. 2dy/g
t = sqrt. 2dy/g
t = sqrt. [2(16 ft)]/32ft/s^2
t = sqrt. 1 s^2
t = 1 s
next, find the initiaL velocity
dx = vix(t)
dx/t = vix
40 ft/1 s = vix
40 ft/s = vix
Thus, the answer is B!
2007-10-29 20:56:42 · answer #2 · answered by Anonymous · 2⤊ 0⤋
e) cannot be solved b/c it doesnt give you the time it took the object to fall form its peek point to the ground.
2007-10-29 16:43:39 · answer #3 · answered by Anonymous · 0⤊ 2⤋