ok, so i have 6 online homework problems that i worked out. i thought that they were all right, but turns out that 2 of them are wrong. i don't know which ones are wrong, and what i did wrong. help please!! i'll write the problems, how i solved, and my answer....i've never used this forum before so i'm not sure if i'm doing this right. appreciate anyone that can help. Question #1: a football is kicked at ground level with a speed of 17.7 m/s at an angle of 46.0 degrees to the horizontal. how much later does it hit the ground. Answer #1: t=2.59 seconds, i found this answer by finding the y component of the vector and using a kinematic equation. Question #2: the pilot of an airplane travelling 1.80E2 km/h wants to drop supplies to flood victims isolated on a patch of land 1.30E2 m below. the supplies should be dropped how many seconds before the plane is directly overhead? (find time in seconds) Answer #2: t=5.15 seconds, i used the kinematic equation....y=y(initial) + v(initial)*time + 1/2 (a) (t^2).......-130=.5(-9.8)t^2 Question #3: William Tell must split the apple atop his son's head from a distance of 26m. when william aims directly at the apple, the arrow is horizontal. at what angle must he aim it to hit the apple if the arrow travels at a speed of 35m/s? Assume that the arrow is released at the same height as the apple. Answer #3: 6.0 degrees, i used the equation sin2x= (Rg)/v(initial)^2 ....R is the horizontal range, which is 26m and g is gravity, v(initial) is 35 Question #4: what average force is required to stop an 1.03E3 kg car in 8.60 seconds if the car is traveling at 96.0 km/h? Answer #4: -3.19E2N....first, i converted 96.0 to m/s (which i'm not sure if i'm supposed to do) then i found the acceleration, then i put that in F=ma Question #5: How much tension must a rope withstand if it is used to accelerate a 2.3E3 kg car vertically upward at .54m/s^2? Answer #5: 23782N (which is 2.4E4 with significant figures) i did FofT=ma-mg Question #6: a person stands on a bathroom scale in a motionless elevator. when the elevator begins to move, the scale briefly reads only 0.77 of the person's regular weight. a) calculate the acceleration of the elevator and b) find the direction of the acceleration...hint (down is negative, up is positive) Answer #6: -2.3 m/s^2 and the direction is down....i did this by looking at an example in the book.....not really sure how to explain it so....i know this is a lot, but if anyone can tell me what i did wrong, it'll make my week. thanks!!!!