# A ball is shot from cannnon into air with upward velocity of 40 ft/sec. The equation that gives the height(h) of the ball at any time id h(t)= -16t^2+40t+1.5. How many seconds rounded to the nearest hundreth will it take the ball to reach the ground?

Mar 13, 2018

$2.56 s$

#### Explanation:

Given equation is $h = - 16 {t}^{2} + 40 t + 1.5$

Put,$t = 0$ in the equation,you will get,$h = 1.5$ that means,the ball was shot from $1.5 f t$ above the ground.

So,when after going upto a maximum height (let,$x$),it reaches the ground,its net displacement will be $x - \left(x + 1.5\right) = - 1.5 f t$(as upward direction is taken positive as per the equation given)

So,if it takes time $t$ then,putting $h = - 1.5$ in the given equation,we get,

$- 1.5 = - 16 {t}^{2} + 40 t + 1.5$

Solving this we get,

$t = 2.56 s$