A ball is thrown from a height of

Question

A ball is thrown from a height of 44

meters with an initial downward velocity of 6/ms

 . The ball's height h

(in meters) after t

seconds is given by the following.=h−44−6t5t2

How long after the ball is thrown does it hit the ground? 

Round your answer(s) to the nearest hundredth.
(If there is more than one answer, use the "or" button.)

Answer 1

This is a physics problem in which a knowledge in parabolic functions can be applied to this problem in which the general distance formula involve the acceleration of gravity is applied to the ball. the general formula is:

 

d(t)=1/2*g*t² + v_o*t + h

 

g = acceleration of gravity = -10 m/s² The negative indicates downward acceleration

v_o= the initial velocity at time equals zero = -6 m/s

h = the initial height from the ground in which the ball is thrown.

 

Now substitute the values in the distance equation in which is:

 

d(t)=1/2*(-10)*t² - 6*t +44 =-5*t² - 6*t + 44 

 

Hence, to find when the ball will hit the ground is when d(t) = 0. This makes the equation being:

 

0 = -5*t² - 6*t + 44

 

Now, solve for t using the quadratic formula. I will leave that up to you.