y = -16t² + 64t +80
The equation above gives the height of an object above the ground, y, in feet, t seconds after it is launched from a platform. How many seconds after it is launched does the object reach the ground?

The correct answer is: 5 seconds

Hint:-
When the object is on ground, its height = y = 0
Step-by-step solution:-
y = -16t² + 64t +80
∴ 0 = -16t² + 64t +80
i.e. 16t² - 64t - 80 = 0
∴ t² - 4t - 5 = 0 …................................................... (Dividing both sides by 16)
Comparing the above quadratic equation with the general form i.e. ax² + bx + c = 0, we get-
a = 1, b = -4 & c = -5
We can solve the above quadratic equation using factorization method as follows-
Steps for Factorization method-
Step-1: a × c = 1 × -5 = -5
Step-2: Factorize -5 in such a way that the 2 factors, when added, gives the value b = -4
i.e. -5 = -5 × 1
& -5 + 1 = -4 = b
∴ t² - 5t + t - 5 = 0
∴ t (t - 5) + 1 (t - 5) = 0
∴ (t + 1) (t - 5) = 0
∴ either t + 1 = 0 or t - 5 = 0
∴ either t = -1 or t = 5
Since time cannot be negative, t = -1 is not a valid option.
∴ t = 5 seconds
Final Answer:-
∴ After 5 seconds of launch, the object would reach the ground.