FutureDoc Posted October 4, 2010 Posted October 4, 2010 So here's a grade 11 physics question my teacher showed to the class with us. John is studying on a cliff 250m high. He throws a rock HORIZONTALLY with an initial velocity of 12.0 m/s (from his height). a) How long did it take the rock to hit the ground below? My teacher showed us on the board: D = V + 1/2 a (t)(squared) then he went on to plugging the numbers into the equation: 250 = 0 + 1/2 (9.8) (t)2 then he got: 7.1s = t I'm confused on how he got 7.1s? Anyone know what he did? I'm worried because there is a homework question similar to what he did. But it has a horizontal component to it, and not a vertical. A bullet is shot horizontally from a gun. If the bullet's speed exiting the muzzle is 325 m/s and the height of the gun above the ground is 2.0m a) how long was the bullet in the air? So from this question I see it is looking for Time. I as well have my velocity which is 325. I'm very confused and stuck on what to do after that. Could anyone try to help me on this one? Sorry if it seems very basic and stupid to you guys. Thanks
Mr Skeptic Posted October 5, 2010 Posted October 5, 2010 Sure. Ignore the horizontal velocity. The vertical velocity is zero.
Echion11 Posted October 12, 2010 Posted October 12, 2010 You already have the equation! D = V + 1/2 a (t)^2 You know the vertical velocity is 0. So V=0. Which leaves you with D= 1/2*a*t^2. You know D (distance)=250 and a=9.81 (depends on where you live). Solving for t does give you 7.14 seconds. Your problem probably is that you don't know that horizontal speed and vertical speed are two seperate things. Your second question is the same as the first.
Genius D'lima Posted October 16, 2010 Posted October 16, 2010 O my, you have no idea how happy I was when i saw this topic, another grade 11 i just joined this forum and i thought it was all university and beyond students and people, but apparently not haha, i guess this post kinda counts as a spam : / well i have physics next sem soooo ya lol.
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