Projectile Automobile


I was recently given a problem on an exam wherein a car fails to make a turn and goes off a cliff as a result. The question then asked what its position was three seconds after going into projectile motion. We were given its initial velocity as 40m/s and that gravity was normal. How does one go about finding position with this information? Thank you for your time.


Cars going off cliffs behave identically with balls thrown from windows. With an initial velocity of 40 m/s in the horizontal direction, in 3 seconds it horizontal location will be 3*40=120 meters from the edge of the cliff.

Its vertical position will be below the cliff edge by the amount which any object falls in 3 seconds under the influence of gravity.

Gravity near the surface of the Earth accelerates objects downward at 9.8 meters per second per second. If you remember the formula for how position depends on time under constant acceleration you can use that. It is d=1.2*a*t2. If t=3 seconds, then t2=3*3=9, so d=1/2*(9.8)*9 or d=44.1 meters below the cliff edge.

If you do not remember the formula, you can get it by recognizing that for a constant acceleration, the average velocity may be assumed for the whole time interval. The average is (start+end)/2. Starting downward velocity was zero so average is just end/2. With constant acceleration the ending velocity is just the acceleration times the time, or in this case 3*9.8. This makes the average velocity in the downward direction 3*9.8/2. To get the final downward position, just multiply the average velocity by the time of 3 seconds, again getting 44.1 meters below the cliff edge.

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