A mountain climber looks up slop and sees an avalanche of boulders accelerating toward her. The boulders are initially 60.0m above, moving at an initial speed of 2.00m/s and accelerating at 4.00 m/s^2 The climber reacts in 0.500s and begins to run down the slope. At what speed must she run in order to reach the bottom of the slope, 30.0m below, before the boulders?

Question

amistre64 · Answer

$$a=\frac{2\Delta x}{t^2}\to\ t=\sqrt{\frac{2\Delta x}{a}}$$
my thought is to time the boulders to the bottom; then see how long the climber

amistre64 · Answer

but im prolly misreading a portion of this; the moment they see the avalanche, its already moving at 2 m/s

anonymous · Answer

2*t +1/2*4*t^2 = 90
(t-.5)V =30

amistre64 · Answer

$$\frac{1}{2}at^2+v_ot+h_o=\Delta x$$
a is downward at 4m/s;  -4
intial height is 60+30=90
and the initial velocity is given as 2ms
$$-2t^2+2t+90=0$$
$$t=\frac{-2\pm\sqrt{4-4(90)(-2)}}{2(-2)}$$the avalanche, if ive done this correctly, might hit the bottom in about 4.73 seconds

anonymous · Answer

~6 seconds...

amistre64 · Answer

i added initial velocity in an up direction dint i :)

anonymous · Answer

yeah:)

anonymous · Answer

or rather...  velocity and acc. have to have the same sign.  I chose 'down' to be '+'   ...for ease.

amistre64 · Answer

i see it better now, thnx ;)