A projectile is shot from the edge of a cliff h = 245 m above ground level with an initial speed of v0 = 145 m/s at an angle of 37.0° with the horizontal, as shown in Fig. 3-39.
(a) Determine the time taken by the projectile to hit point P at ground level.
s
(b) Determine the range X of the projectile as measured from the base of the cliff.
km
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity (take up and to the right as positive directions).
I got 115.8 m/s (horizontal component)
I got 87.26 m/s (vertical component)-is this right?
(d) What is the the magnitude of the velocity?
m/s
(e) What is the angle made by the velocity vector with the horizontal?
° (below the horizontal)
I don't get any of this. I don't know what they mean by range. Is magnitude the same as velosity?I don't understand what they are asking me to do.
2006-10-24 15:11:28 · 4 answers · asked by beast 1 in Science & Mathematics ➔ Engineering
With Fig. 3-39 we might have a clue
2006-10-24 15:19:56 · answer #1 · answered by myothernewname 6 · 0⤊ 0⤋
Range means distance
The magnitude of the velocity is the VALUE independent of direction.
Decompose the velocity into vertical and horizontal components. The vertical component provides the vertical kinetic energy which must equal the potential energy gain at the top of the flight.
2006-10-24 15:37:06 · answer #2 · answered by arbiter007 6 · 1⤊ 0⤋
Might want to consider switching majors if you don't get this.
2006-10-25 05:04:05 · answer #3 · answered by Jeffrey S 6 · 0⤊ 0⤋
save yourself some time, bomb the test
2006-10-24 15:21:34 · answer #4 · answered by Anonymous · 0⤊ 0⤋