Question

For this exercise, use the position function s(t) = −4.9t2 + 250, which gives the height (in meters) of an object that has fallen for t seconds from a height of 250 meters. The velocity at time t = a seconds is given by lim t→a s(a) − s(t) a − t . When will the object hit the ground? At what velocity, v, will the object impact the ground?

Answer 1

Answer:

t=7.14s

v=-69.972 m/s

Explanation:

Position function

$s(t)=-4.9*t^{2}+250$

Velocity is the derivative of position function

$V(t)=\frac{dx}{dt}\\V(t)=-2*4.9*t\\V(t)=-9.8*t$

The time the object hit the ground can be find by the given function know that the position is going to be 0m

$s(t)=-4.9*t^{2}+250$

$s(t)=0\\0=-4.9*t^{2} +250\\t=\sqrt{\frac{250}{4.9}}\\t=7.14s$

Check:

$s(7.14)=-4.9*(7.14s)^{2}+250\\ s(7.14)=-250+250\\s(7.14)=0m$

So the velocity can be find using the time discovery before and using the same function but with the derivate

$V(t)=-2*4.9*t\\V(7.14)=-2*4.9*(7.14)\\V(7.14)=-69.972 \frac{m}{s}$

The velocity is negative because the object is moving downward