EK Physics Book Q17

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EK Physics Book Q17

This seems like an easy question but for some reason, I don't get it and even after reading the solution explanation, I am confused. Can someone explain this?

Here is my reasoning:

It goes from 100m to 80m in 2 sec. Displacement is 20 meter in 2 second and in 4 seconds, displacement should be 40 m, so 100m (initial position) - 40 m = 60 m (new position), answer choice A.

I also tried using the d = v1t + 1/2at^2 equation

d = (10)(4) + 1/2 (-10)(4)^2 = - 40m

Again, 100 m - 40 m = 60 m (answer choice A)

 

[monkeyvokes]

can you post the question?

 

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17. If an apple that is dropped from an altitude of 100 m
reaches an altitude of 80 m after falling for t = 2 seconds.
what altitude will it be at in t = 4 seconds?

A. 60m
B. 40m
C. 20 m
D. Om

 

[monkeyvokes]

x = 1/2 * a * T^2

in the first 2 seconds at 10m/s^2 this apple covers 20 meters.

now try putting in 4 seconds instead of two seconds.. the apple covers 80 meters. The apple's new altitude should be 20meters

 

[monkeyvokes]

EK Physics Book Q17


I also tried using the d = v1t + 1/2at^2 equation

d = (10)(4) = 1/2 (-10)(4)^2 = - 40m

1/2 * 10 * 16 isn't 40 (unless i'm typing this in wrong)