Question

An object is launched at an initial velocity of 19.6 meters per second from an initial height of 24.5 meters. Which equation can be used to find the number of seconds it takes the object to hit the ground?

Answer 1

the equation is:
d= vi×t + 1/2 a×t^2

Answer 2

Answer:

Quadratic equation is : $4.9t^2-19.6t+24.5=0$

Step-by-step explanation:

It is given that,

Initial velocity of an object, u = 19.6 m/s

Initial height, s = 24.5 m

We have to write the equation to find the number of seconds it takes the object to hit the ground. It can be calculated using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$..........(1)

s is displacement of object

u is initial velocity of an object

g is acceleration due to gravity

t is time taken  in seconds      

Here, s = h and a = g = -9.8 m/s²

So, equation (1) becomes :

$24.5=19.6t-\dfrac{1}{2}\times 9.8t^2$

$4.9t^2-19.6t+24.5=0$

Hence, the above quadratic equation is used to find the time taken by the object to hit the ground.