a cannonball is fired from the top of the hill at the angle of 40 degrees. The hill is 301m high. If the cannonball lands at 399 m away from the base of the hill. What is the initial velocity of the cannonball?
2006-10-30 15:48:48 · 2 answers · asked by Ha!! 2 in Science & Mathematics ➔ Physics
Split the problem into the X components (how far the cannon ball travels horizontally - 399m) and the y component (how far the cannon ball travels vertically - 301m). Then you'll use the equation: X=VT+(1/2)AT^2 where x is the distance, v is velocity, t is time, and a is acceleration. Use that equation independently for each component of the problem so you end up with:
Vcos(40)T=399 (since the acceleration in the horizontal direction doesnt change and the horizontal velocity is represented by the cos(40)*V.
and for the Y component:
301=Vsin(40)T+1/2 * -9.8 * T^2
since acceleration due to gravity is -9.8 and the initial velocity in the y direction is Vsin(40).
since you don't want T to be in the equation take one and solve for T in the second equation using the equation (-b+sqrt(b^2-4ac))/(2a) where the equation is in the format of aT^2+bT+C. now that you have T in terms of V, plug that into the first equation and then solve for V.
thats a pretty bad explanation but its harder to type than it is to write when you can show a picture to better illustrate why things are like they are.
2006-10-30 16:10:31 · answer #1 · answered by Cole 2 · 1⤊ 0⤋
Is it 40 dgrees above the horizon, or 40 below? Is this "hill" really a cliff? Is the ball fired from the top of a cliff, and lands 399 from the base of the cliff? Who gave you this question?
2006-10-31 00:03:23 · answer #2 · answered by ? 6 · 0⤊ 0⤋