An object is launched into the air. The height (in meters) above the ground is given by the expression h= 50t - 5t^2.

Question

During which time is its height above the ground higher than 80 meters?

Answer 1

h= 50t - 5t^2

We have to calculate time when h=80, on substituting the value of h in above equation we get:

80 = 50t - 5t^2

or  t^2 - 10t + 16 =0

so we get, t = 2 or 8

So at time t = 2 seconds or 8 seconds the ball will be at the height of 80 m

2 seconds when the ball is going up.

8 seconds when the ball is falling down.

Comments

Thank you

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