a rock is thrown upward from the edge of a 50 m cliff overlooking a lake, with an initial velocity of 17.2 m/s. Define variables and write and equation that models the height of the rock.

Question

anonymous · Answer

we will consider only the vertical direction

use the equation:

$$y_2(t) - y_1 = V_0 t + \frac{ 1 }{ 2 }a t^2$$
where
y2 and y1 are the heights at a time 2 and a time 1
V_0 is the initial veloity
a is acceleration [gravity in our case = -9.8 m/s^2]
t is the difference in time between height 1 and 2

anonymous · Answer

in our model, y_1 is the initial height [50], so simplified, our model is:

$$y_2(t)=50+17.2t−4.4t^2$$

anonymous · Answer

where does the 4.4t^2 come from?

anonymous · Answer

wow, made a mistake.

should be -9.8/2 = 4.9

anonymous · Answer

$$y_2(t)=50+17.2t−4.9t^2$$

anonymous · Answer

where does the -9.8 come from?

anonymous · Answer

gravity = -9.8 m/s^2  = acceleration on rock

anonymous · Answer

oh ok thanks!