You accidentally throw your car keys horizontally at 7.0 m/s from a cliff 72 m high. How far from the base of?

Question

You accidentally throw your car keys horizontally at 7.0 m/s from a cliff 72 m high. How far from the base of the cliff should you look for the keys?

Answer 1

If t = time to reach the base,
72 = (4.9) t^2
=> t = √(72)/(4.9) = 3.833 s

At constant horizontal velocity of 7.0 m/s, its distance from the base will be
7.0*t = (7.0) * (3.833) = 26.8 m.

Initially, I had made calculation error which I corrected after seeing gebobs's answer.

Answer 2

question takes to steps. to discover how a methods they finally end up from the element they are thrown, you would be able to desire to be attentive to how long will or not that's until they strike the floor. so: equation of action is a million/2at^2 + vt + x = 0 at time 0, the downward velocity is 0 so the midsection term will become 0. X is the area up the cliff. Equation will become: a million/2at^2 + 0 + seventy six = 0 rearrange the words: at^2 = -76m * 2 at^2 = -152m Acceleration by way of gravity is -9.8m/sec^2 so: t^2 = -152/-9.8 t^2 = 15.fifty one take sqrt of the two aspects t = 3.938 sec understanding that the keys are interior the air for that long the area from the cliff is merely velocity * time 5 m/sec * 3.ninety 3 sec = 19.69m

Answer 3

I'd slap myself for doing this and then catch a taxi. Cliff's are dangerous places. Good luck to you.

Answer 4

Time to fall:

t = sqrt(2x/g) = sqrt(2*72/9.8) = 3.8 sec

Distance from base:

x = vt = 7.0 * 3.8 = 27m

All answers to two sig figs