Harrison plays a game in which he uses a net to catch balls that are randomly launched from holes in the ground. the height of one ball after t seconds is represented by the function f(t) = –16t2 + 20t. after how many seconds does the ball return to the ground? round to the hundredths place as needed. edit: answer is 1.25

⇒ t = 1.25 seconds  

Step-by-step explanation:

The height of one ball after t seconds is represented by the function f(t) = -16·t² + 20·t

We need to find : After how many seconds does the ball return to the ground

Solution :

When the ball will return to the ground the height of the ball will be 0

⇒ f(t) = 0

So, finding the value of t when f(t) = 0

⇒ -16·t² + 20·t = 0

⇒ t·( -16 t + 20 ) = 0

⇒ t = 0 or ( -16 t + 20 ) = 0

t = 0 is rejected because initially the ball is not on the ground.

⇒ ( -16 t + 20 ) = 0

⇒ - 16t = -20

⇒ 16t = 20

⇒ t = 1.25 seconds

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.

step-by-step explanation: