Question

18. A tennis ball is launched straight upward with an initial velocity of 24.5 m/s from the edge of a cliff that is 117.6 meters above the ground. Which quadratic equation could be used to correctly determine when the ball will hit the ground:4.9t2 + 24.5t + 117.6 = 0-4.9t2 - 24.5t + 117.6 = 0-4.9t2 + 24.5t - 117.6 = 04.9t2 + 24.5t - 117.6 = 0-4.9t2 + 24.5t + 117.6 = 0

Answer 1

Take into account that in general, the motion of an object with an initial velocity pointing upward, but under the influence of the gravitational force (which eventually will makes that the object come back to the ground) is given by:

`y=y_o+v_ot-(1)/(2)gt^2`

In this case:

yo = 117.6 m

vo = 24.5 m/s

g = 9.8 m/s^2

By replacing the previous values without physical units, into the expression for the height of the tennid ball, you have:

`\begin{gathered} y=117.6+24.5t-(9.8)/(2)t^2 \\ y=117.6+24.5t-4.9t^2 \end{gathered}`

Now, consider that we are interested in the situation at wich the tennis ball hit the ground, which is the same that to sat y = 0. Then, you can write:

`0=117.6+24.5t-4.9t^2`

By ordering the previous expression you obtain:

`-4.9t^2+24.5t+117.6=0`

The previous result is the required equation.