`y=-5t^(2)+2t+20`
x=2t
A cannonball is shot out of a cannon at a `45^(@)` angle with an approxiate speed of 283m/s. The cannon from which the ball was fired sits on the edges of a cliff, and its height above the ground is 20 meters. The equations given above represent the cannonball's height above the ground(y) and its horizontal distance (x) from the face of the cliff, t seconds after the ball was fired, where `tge0`. How many seconds after the ball was fired does its vertical height above the ground equal its horizontal distance from the cliff?

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

B

Explanation

Difficult: Medium
Category: Passport to Advanced Math/ Quadratics
Strategy Advice: In real-world scenario's a good bit of the text is probably just providing the context of the question. Be sure to focus on what the question is asking (here, find t when x and y are equal).
Getting to the Answer: Because both equations are given in terms of t, set the the equation equal to each other and simplify:
`-5t^(2)+2t+20=0`
`-5t^(2)+20=0`
The resulting equation is quadratic, so you can either factor and solve, or use square rooting.

Time can't negative, so t must equal 2. This matches (B).