Answer to Question #280181 in Calculus for ash

Question

A ball is dropped from a height of 126.8 meters. Its height above ground (in meters) t seconds later is given by: s(t)=−17t^2 +126.8

What is the instantaneous velocity of the ball when it strikes the ground?

Answer 1

$$v(t)=s'(t)=-34t$$

$$s(t)=0=>−17t^2 +126.8=0, t\geq0$$

$$t=\sqrt{126.8/17}$$

$$v=-34\sqrt{126.8/17}\approx-92.857\ m/s$$

The instantaneous velocity of the ball when it strikes the ground is

$-92.857\ m/s$ (downward speed with value of $92.857\ m/s$ ).