To solve this problem, we will set up the equation y = -4.9t^2 + 9, where y represents the height of the rock above the crevasse and t represents the time in seconds.
Next, we will set the height y to -2 meters (2 meters below ground level) and solve for t:
-2 = -4.9t^2 + 9
-4.9t^2 = 11
t^2 = 11 / 4.9
t^2 ≈ 2.2449
t ≈ √2.2449
t ≈ 1.5 seconds (rounded to the nearest tenth of a second)
Therefore, the rock will hit the bottom of the crevasse 2 meters below ground level in approximately 1.5 seconds.
While hiking, Julia kicked a small rock off a ledge that was meters above a crevasse. Use desmos or GeoGebra to graph the situation using the formula y= -4.9t2 + 9. Use the graph to determine when the rock will hit the bottom of the crevasse, 2 meters below ground level. Round your answer to the nearest tenth of a second. (1 point)
The rock will hit the bottom of the crevasse in seconds.
1 answer
1 answer