Question

PLEASE HELP ASAP

A projectile is shot horizontally off a cliff at a speed of 27 m/s. It hits the ground
2.2 seconds later.
How high is the cliff, in meters?

Answer 1

Approximately
`11\; {\rm m}`, assuming that gravitational acceleration is
`g = 9.81\; {\rm m\cdot s^(-2)}` and that air resistance is negligible.

Step-by-step explanation:

The question is asking for the vertical displacement,
`x_(y)`.

Under the assumptions, acceleration in the vertical direction would be
`a_(y) = -g = (-9.81)\; {\rm m\cdot s^(-2)}`.

Since the projectile is launched horizontally, initial vertical velocity would be zero:
`u_(y) = 0\; {\rm m\cdot s^(-1)}`.

It is given that the projectile is in this motion for
`t = 2.2\; {\rm s}`. Apply the SUVAT equation
`x = (1/2)\, a\, t^(2) + u\, t` and solve for the vertical displacement:

`\begin{aligned}x_(y) &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t \\ &= (1)/(2)\, ((-9.81)\; {\rm m\cdot s^(-2)})\, (2.2\; {\rm s})^(2) + (0\; {\rm m\cdot s^(-1)})\, (2.2\; {\rm s}) \\ &= ((-9.81)\, (2.2)^(2))/(2)\; {\rm m} \\ &\approx (-11)\; {\rm m} \end{aligned}`.

Note that the value of vertical displacement is negative since the projectile landed below where it was launched. The height of this cliff would be approximately
`11\; {\rm m}`.