I have an exam and I don't get this Physics problem.?

Question

Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The speed of A is 2.50 m/s and the speed of B is 3.40 m/s. Both grenades remain in air for 3.77 s. Assume that the acceleration due to gravity is 9.86 m/s squared. What is the distance between A and B if they are thrown along the same straight line?


P.S. Please don't tell me to 'do my own hw.' If you don't want to help, don't. But don't say that cuz I would obviously do it if I understood it. Thanks.

Answer 1

In order to solve this problem one must find out what the horizontal component of each grenade's velocity as it leaves the cliff.
We are told each grenade's launch velocity, but not at which angle they are thrown.

Using the equation,
d = 1/2 at^2
Where d is the distance the grenades fall, a is the gravitational acceleration, and t is the fall time of the grenades (what it should be without any additional, initial, vertical velocity), we can find out if the grenades were thrown at an angle, or if they were thrown horizontally.
Plugging it, we find that the free fall time for the grenades to hit the ground if they were thrown horizontally would be 3.77 seconds...the same time given in the problem, so in this case, each grenade's initial velocity was also its horizontal velocity.

Since gravity only acting on the vertical direction, the grenades' horizontal velocity throughout the fall is constant.

To find the horizontal distance separating the grenades when they hit the ground, find the difference in their velocities, then multiply by the time they are traveling.

Grenade A is 2.5 m/s, B is 3.4 m/s ... a difference of 0.9 m/s and they fall for 3.77 seconds. By the time the two grenades hit the ground they should be 3.4 meters apart (B ahead of A).

Answer 2

Since both grenades were thrown horizontally, and are accelerating downwards with the same acceleration, they always remain separated by some horizontal length.

(The vertical information is irrelevant except for checking its consisitency; to the number of figures quoted, they're consistent.)

In addition to being both thrown out horizontally, they were in fact thrown IN THE SAME DIRECTION. Then, to find the distance between them after 3.77 secs, you need only the DIFFERENCE in their horizontal speeds, which is 0.90 m/s.

Thus, after 3.77 s, they're separated by 0.90 m/s x 3.77 s = 3.4m.

(Note that you wouldn't be justified in saying 3.39 m, worse yet 3.393 m; the separate speeds of A and B are given only to 3 sig. figs., so you only know their DIFFERENCE --- the data required to determine your answer --- to 2 sig. figs. It's all too easy to forget such niceties. Note also that in GARY R's response, all the figures beyond ".8" in his 12.818 m for B's flight distance are unreliable; starting by multiplying two 3-sig.-fig. numbers CANNOT give you a result accurate to 5 sig. figs! So the apparent increase in significance by subtracting his A and B flight distances from one another is simply spurious. Sorry about it, but that's the way the world is.)

I hope this helped. Live long and prosper.

Answer 3

It is almost a trick question. Verticle distance doesn't matter here. The first will go 2.50m/s x 3.77s = 9.425 meters, the second will go 3.40m/s x 3.77s = 12.818 meters.

The 70 meters high and the 9.86 will give you the 3.77 seconds.

time = sqrt(2h/g)

Answer 4

F = ma. A is the replace in speed so in case you concentration on that at time t = 0 the ball change into nonetheless and speed v = 0 then at t=a million the guy hit it and gave it a speed of sixty 3 m/s then the acceleration a might want to be sixty 3 m/s². You don’t understand F = ma? quite?? reliable luck the following day!!

Answer 5

equation for free fall

y = y(initial) + v(initial)xtime - (.5)gt^2

where y(initial) is the initial height, v(initial) the initial velocity, g is the acceleration due to gravity, an t is time. Plug your info there and you got your answer.