Question

Projectile Motion Problems

1. A movie scene has a car drive off a cliff.
a. If the car took 5.5 s to reach the ground, how high was the cliff?
b. If the car had an initial velocity of 26 m/s, how far from the cliff bottom did the car
land?

Answer 1

Height of the cliff would be approximately
`150\; {\rm m}`.

The landing site would be approximately
`143\; {\rm m}` from the bottom of the cliff.

(Assume that
`g = 9.81\; {\rm m\cdot s^(-2)}`, air resistance is negligible, the top of the cliff is level, and that the cliff is vertical.)

Step-by-step explanation:

Assume that air resistance is negligible. The vertical acceleration of the vehicle would be constantly
`a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` during the fall.

If the top of the cliff is level, initial vertical velocity
`u_(y)` would be
`0\; {\rm m\cdot s^(-1)}`.

Apply the SUVAT equation to find the vertical displacement
`x_(y)` of the vehicle in that
`t = 5.5\; {\rm s}`.

`\begin{aligned}x_(y) &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t \\ &= (1)/(2)\, (-9.81)(5.5)^(2) \; {\rm m} + (0)\, (5.5)\; {\rm m} \\ &\approx (-150)\; {\rm m}\end{aligned}`.

In other words, the vehicle landed approximately
`150\; {\rm m}` below where it took off. The height of the cliff would be
`150\; {\rm m}\!`.

Also under the assumption that air resistance is negligible, the horizontal velocity of the vehicle would be constant:
`v_(x) = 26\; {\rm m\cdot s^(-1)}`.

Since horizontal velocity is constant, multiply this velocity by by time to find the horizontal displacement
`x_(x)`:

`\begin{aligned}x_(x) &= v_(x)\, t \\ &= (26)\, (5.5)\; {\rm m} \\ &= 143\; {\rm m} \end{aligned}`.