Question

A ball is thrown up in the air from the top of a building 6 meters high. Its hight
above the ground after t seconds is f(t) = 6+13-t5t^2.

Find the velocity of the ball when it hits the ground?

Answer 1

The equation is f(t) = 6 + 13t - 5t^2.

When ball hits the ground, height becomes zero.

6 + 13t - 5t^2 = 0

5t^2 - 13t - 6= 0

t = [-(-13) + sqrt{(-13)^2-4(5)(-6)}] / 2(5)

t = 3, -0.39

Time cannot be negative. so t = 3 seconds.

Find the velocity by differentiating the function with respect to t.

f'(t) = 13 - 10t.

Substitute t = 3 in the above equation.

f'(t) = 13 - 10(3) = - 17 m/sec.

Negative sign represents ball is falling.