an airplane , diving at an angel of 53 degrees with the vertical, relases a projectile at an altitude of 730 meters. the projectile hits the ground 5 seconds after being released.
(a) what is the speed of the aircraft?
(b) how far did the projectile travel horizontally during its flight?
what were the...
(c) horizontal and (d) vertical components of its velocity just before striking the ground?...
plzzz!! any help.. i'm in desperate need of help... I don't want answers... just a way to solve it.. or hints.. plzz.. thank you!!
2006-10-16 19:58:24 · 4 answers · asked by betodemise46 3 in Science & Mathematics ➔ Physics
The projectile will have two components of velocity. vertical and horizontal. Only the vertical component determines how long before it reaches the ground. Let vy be that vertical component of initial velocity. The speed of the projectile toward the ground is vy + g*t, where g = accel of gravity = 9.8m/sec^2. The distance it travels vertically is ∫vdt = ∫(vy + g*t)dt = vy*t + .5*g*t^2; you know t to reach ground, 730m down; so 730 = vy*5 + .5*g*25; solve for vy. vy is the vertical component of the velocity vector of the airplane, which is at 53º from the horizontal. Draw a vector diagram showing vy and the angle to get the plane's velocity.
The horizontal velocity of the projectile (ignoring air friction) does not change; you can find the horizontal component from the same vector diagram as above. The projectile will travel at that velocity until it hits the ground t=5 seconds later.
As for the velocity components at reaching ground, as said above, the horizontal velocity stays the same. The vertical velocity formula I gave you above (vy + g*t).
I hope this helps.
2006-10-16 20:37:33 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋
Let the components of the projectile velocity at the release point be vx and vy. The angle is given, so once vx and vy are known, the aircraft speed can be found by trigonometry. The projectile falls for 5 seconds, with its initial vertical velocity vy augmented by the force of gravity. In 5 seconds, the projectile will have picked up an additional 49 m/sec of velocity. The average velocity is vy+24.5 m/sec, so 730 divided by this = 5 seconds. Use sin 53 = 0.8, and cos 53 = 0.6. There is enough stuff here to give you the solution if you put it together.
2006-10-17 13:27:13 · answer #2 · answered by Anonymous · 1⤊ 0⤋
i) calculate the speed of the projectile. 730/5
ii) break the speed in horizontal and vertical component(draw)
iii) use tan to find the components and find the range
iv) use the formula to find the acceleration.
v) the horizontal component component will be the same because it will not be affected by gravity.
2006-10-17 03:43:17 · answer #3 · answered by IQEinsten 2 · 1⤊ 0⤋
Try this web site:
http://www.physicsclassroom.com/Class/vectors/U3L2a.html
2006-10-17 03:17:48 · answer #4 · answered by jm 2 · 1⤊ 0⤋