Question

Ben throws the ball to Lana. After T seconds of throwing, the height of the ball (m)

from the ground surface, the function h(t) = -1.7t² + 4,6t + 1,8

a) How high is the ball 2 seconds after the throw? 

b) Lana can't catch the ball. How long before the ball hits the ground? 

c) At what time is the ball peaking? What is the altitude of the ball in this case?

Answer 1

a)

$$h(2) = -1.7(2)^2 + 4.6(2) + 1.8$$

$$h(2) =4.2\ m$$

b)

$$h(t)=0=> -1.7t^2 + 4.6t + 1.8=0, t\geq0$$

$$D=(4.6)^2-4(-1.7)(1.8)=33.4$$

$$t=\dfrac{-4.6\pm\sqrt{33.4}}{2(-1.7)}=\dfrac{2.3\pm\sqrt{8.35}}{1.7}$$

Since $t\geq0,$ we take

$$t=\dfrac{2.3+\sqrt{8.35}}{1.7}\ sec$$

$$t\approx3.053\ sec$$

c)

$$t_v=-\dfrac{4.6}{2(-1.7)}$$

$$t_v=\dfrac{23}{17}sec$$

$$t_v\approx1.353\ sec$$

$$h(t_v) = -1.7(\dfrac{23}{17})^2 + 4.6(\dfrac{23}{17}) + 1.8$$

$$h(t_v)\approx4.912\ m$$