Question

An object is launched at 9.8 meters per second from 73.5 meter tall platform. The objects height s (in meters) after t seconds I’d given by the equation s(t)= -4.9t2-9.8+73.5. Use the factoring to determine when the object hits the ground

Answer 1

Answer:

The object hits the ground in 5 seconds.

Step-by-step explanation:

Given : An object is launched at 9.8 meters per second from 73.5 meter tall platform. The objects height s (in meters) after t seconds I’d given by the equation $s(t)= -4.9t^2-9.8t+73.5$.

To find : Use the factoring to determine when the object hits the ground ?

Solution :

When the object hits the ground the distance is zero.

i.e. s(t)=0

So, $-4.9t^2-9.8t+73.5=0$

Applying quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=-4.9, b=-9.8 and c=73.5

$t=\frac{-(-9.8)\pm\sqrt{(-9.8)^2-4(-4.9)(73.5)}}{2(-4.9)}$

$t=\frac{9.8\pm\sqrt{1536.64}}{-9.8}$

$t=\frac{9.8\pm 39.2}{-9.8}$

$t=1\pm 4$

$t=1+4,1-4$

$t=5,-3$

Reject t=-3

Therefore, the object hits the ground in 5 seconds.