Question

A ball is thrown from a height of 45 meters with an initial downward velocity of 10 m/s. The ball's height h (in meters) after t seconds is given by the following. h=45-10 t- 5 t^2. How long after the ball is thrown does it hit the ground?

Answer 1

Answer:

2.16 s

Step-by-step explanation:

When it hits the ground, height h=0 hence

$0=45-10t-5t^{2}$

$5t^{2}+10t-45=0$

$t^{2}+2t-9=0$

Using quadratic formula

$x=-1+\sqrt 10\\ x=-1-\sqrt 10$

Since time can't be negative, $-1+\sqrt 10$

t=2.16 s