- #1

reyrey389

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This was originally posted in a non-homework forum so it does not use the template.

A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.998 s. How much time passes during his entire trip from the top down to the water? Ignore air resistance.

Attempt:

t = 0.998 approx 1 sec

vo = 0 m/s (v initial)

g = -9.8 m/s^2

I solved for vf ( v final) using vf = -g * t = -9.8 m/s. This is also the initial velocity of falling the second half.

Now using v_f, v_o, and g, I solved for displacement. using vf^2 - vo^2 = 2a*Delta x

Thus the half the distance would be 4.9 meters. Now I solved for the time to fall the second half of the height using

vo = =-9.8 m/s , a = -9.8, h = 4.9 meters.

-4.9 = - 9.8t - 4.9t^2

t^2 + 2t = 1

t^2 + 2t - 1 = 1 + 1

(t-1)^2 = 2 so t = sqrt(2) - 1 or 0.41 seconds.

For t_total I did 0.41 + 1 second from the given to get 1.41 seconds.

I'm not sure if this is correct when I searched on yahoo answers, the answer reported was 1.13, but my work seems to make sense. If anyone can please elaborate, I'd appreciate

Attempt:

t = 0.998 approx 1 sec

vo = 0 m/s (v initial)

g = -9.8 m/s^2

I solved for vf ( v final) using vf = -g * t = -9.8 m/s. This is also the initial velocity of falling the second half.

Now using v_f, v_o, and g, I solved for displacement. using vf^2 - vo^2 = 2a*Delta x

Thus the half the distance would be 4.9 meters. Now I solved for the time to fall the second half of the height using

vo = =-9.8 m/s , a = -9.8, h = 4.9 meters.

-4.9 = - 9.8t - 4.9t^2

t^2 + 2t = 1

t^2 + 2t - 1 = 1 + 1

(t-1)^2 = 2 so t = sqrt(2) - 1 or 0.41 seconds.

For t_total I did 0.41 + 1 second from the given to get 1.41 seconds.

I'm not sure if this is correct when I searched on yahoo answers, the answer reported was 1.13, but my work seems to make sense. If anyone can please elaborate, I'd appreciate