Functions

[LaryssaLacerda]

A motorcycle stuntwoman is planning a daring jump across a 352-foot-wide river
and sets up identical ramps on both sides of the river. Her vertical speed on
take-off is 50 feet per second, so her height(in feet) above the ramp at the
time t(in seconds, after take-off) will be described by the function
h(t)=50t-16t^2. She has determined that if her horizontal speed at the time of
take-off is 124 feet per second, she will land exactly on top of the ramp on the
other side. Her horizontal speed will remain constant.

Is she correct? If not, will she fall into the river or overshoot the landing?

WORK DONE :

I have not idea on how to even start this problem.. I am totally lost..

Please help me.. Thank-you so much.

 

[soroban]

Hello, Laryssa!

A motorcycle stuntwoman is planning a daring jump across a 352-foot-wide river
and sets up identical ramps on both sides of the river.
Her height (in feet) above the ramp at time \(\displaystyle t\) is: \(\displaystyle \,h(t)\:=\:50t\,-\,16t^2\)
She has determined that if her horizontal speed at the time of take-off is 124 ft/sec,
she will land exactly on top of the ramp on the other side.
Her horizontal speed will remain constant.

Is she correct? .If not, will she fall into the river or overshoot the landing?

At take-off and landing, her height is zero.
. . Hence: \(\displaystyle \,50t\,-\,16t^2\:=\:0\;\;\Rightarrow\;\;t\:=\:0,\,\frac{25}{8}\)
Hence, she takes off at \(\displaystyle t\,=\,0\) and is airborne for \(\displaystyle 3\frac{1}{8}\) seconds.

Her horizontal speed is a constant 124 ft/sec.
In \(\displaystyle t\) seconds, she will travel \(\displaystyle 124t\) feet horizontally.

In \(\displaystyle 3\frac{1}{8}\) seconds, her horizontal distance will be: \(\displaystyle \,124\left(\frac{25}{8}\right)\:=\:387.5\) feet.

Therefore, she will overshoot the landing by: \(\displaystyle \,387.5\,-\,352\:=\:35.5\) feet.