Projectile Motion?

Question

During a volcanic eruption, chucks of solid rock can be blasted out of the volcano; these projectiles are called volcanic blocks. Rock is blasted at an elevation of 3300 meters at an angle of 35 degrees.

a) At what initial velocity would a block have to be to be ejected, at 35 degrees to the horizontal, from the vent to the foot of the volcano 9400 meters away?

b) What is the time of flight?

I can do B, but I dunno how to find initial velocity, most of the problems I did with projectile motion already gave me initial velocity. Here's what I did:

I used the formula y-y(0)=vt-1/2at^2. I'm going to assume the final velocity of this whole thing is 0, and gravity is -9.8m/s^2, so:

-3300=-4.9t^2; t=26

I used that "t" to plug into: v=v(0)+AT

0=v(0)-9.8(26); V(0)=255 m/s. The answer that the book got was 260 m/s, can anyone show me what I did wrong and how to do the problem? Thanks!

Answer 1

This problem is more complex than you have assumed. The rock is blasted at 35º to the horizontal, and therefore will reach a height more than the volcano's 3300m. Consider the trajectory in three segments with times t1 t2 and t3. t1 is the time to reach peak altitude, t2 is the time to fall to the original altitude (volcano elevation), and t3 is the time to fall to the foot of the volcano. The horizontal component of the block's velocity is then the horizontal distance traveled (9400m) divided by the total time t1+t2+t3.

The times can be calculated in terms of the initial vertical velocity component v0. t1 = v0/g. t1 = t2 by symmetry. t3 is found from the fact that the downward velocity at the start of t3 is equal to v0 (again by symmetry), and the block accelerates downward, so the distance traveled is given by v0*t3 + 0.5*g*t3^2, and this equals the elevation of the mountain.

The vertical and horizontal velocity components are related by the tangent of the initial angle

v0/vh = tan(35º)

These formulas can be combined to solve for either v0 or vh. Then the initial velocity is v0/sin35º or vh/cos35º; then use v0 to compute t1, t2 and t3 to get the total time of flight.

EDIT: I have worked out this problem using a little simpler approach than above. I also get 255.5 m/sec. I don't see how you came up with that answer from your description, but you can find my solution in two pages here:

http://img407.imageshack.us/img407/5928/volcano1ik5.png
http://img404.imageshack.us/img404/2774/volcano2ct9.png

Answer 2

The destructive sign refers back to the path of the rigidity, that's downward. by ability of convention, destructive velocity and destructive acceleration are downward. in case you place g as helpful, then a projectile released upwards could have destructive velocity, which does no longer make intuitive experience.

Answer 3

take ,initial velocity = u
horizontal distance = x
vertical distance =y
agle of projection =Q
NOW,
FOR HORIZONTAL COMPONENTS:
u*cosQ *t = x
so, t = x/(u*cosQ)
FOR VERTICAL COMPONENTS :
y = u*sinQ *t -0.5*g*t*t
so,y = u*sineQ *x/(u*cosQ)-0.5*g*[x/(u*cosQ)]*[x/(u*cosQ)]
COMPUTING OUT THIS,
u = square root of [(g*x*x)/{2*cosQ*cosQ(x*tanQ-y)}]
so,u =