Heidi T. answered • 09/09/19
MS in Mathematics, PhD in Physics, 7+ years teaching experience
Anytime you have motion under gravity, the time of fall depends ONLY on the fall distance in the y-direction. This seems a bit counter-intuitive, but an object dropped and an object propelled horizontally with any velocity will hit the ground at the same time.
Before starting any question, draw a picture and write everything known on it. Define your coordinate system. This can be done in any way, but choose the way that simplifies the math in the problem.
In this case, define the origin of the coordinate system to be the location from where the rock is thrown, positive x in the direction the rock is thrown, and positive y downward.
The general equation of motion is y = y0 + v0y *t + (1/2) * g * t^2 Both y0 and v0 are zero - yo because of the definition of the origin of the coordinate system, v0 because the initial velocity is in the horizontal (x-direction)
The equation of motion simplifies to y = (1/2) * g * t^2. Solving for t,
t = sqrt[ 2*y / g]
y = 108 m (given), g = 9.8 m/s^2 (acceleration due to gravity - if you are using g = 10 m/s in your class, will need to recalculate the answer)
t = sqrt[ 2*(108 m) / (9.8 m/s^2)] = 9.95 s = 10 s to two significant figures.
