The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where should a target be placed so that the toy car lands on it?

Answer 1

Hi Maisey P.,

Not enough restriction to the problem to answer. It would be necessary to know the angle at which the car was launched with respect to the plane of the ground. After that, you can assign vectors to the x- and y-components of the car's velocity. Next, solve for y(t), using vy(0), the initial position y(0) = +0.5 m, and the gravity acceleration (-9.8 m s^(-2). IF vy(0) is 0, then y(t) = 0.5 - 4.9*t^2, and ~0.319 s suffices to reach the ground. The car traverses 5 m s^(-1)*0.319 s horizontally, or ~1.597 m, in that time interval. However, since the initial conditions were only given to 1 significant digit, you would be justified, even compelled, in rounding to 2 m, even though that would potentially cause the car to not land on the target at all. That issue arises anytime a car flies through the air for any reason, so don't drive faster than a road permits. Or even close to the limit, since your traction will suffer even if you don't actually go airborne.

-- Cheers, --Mr. d.