103
A ball is thrown in the air so that $t$ seconds after it is thrown, its height $h$ metre above its starting point is given by the polynomial $h = 25t - 5t^2$.
Observe the graph of the polynomial and answer the following questions :
(i) Write zeroes of the given polynomial.
(ii) Find the maximum height achieved by ball.
(iii) (a) After throwing upward, how much time did the ball take to reach to the height of $30$ m?
OR
(iii) (b) Find the two different values of $t$ when the height of the ball was $20$ m.
Observe the graph of the polynomial and answer the following questions :
(i) Write zeroes of the given polynomial.
(ii) Find the maximum height achieved by ball.
(iii) (a) After throwing upward, how much time did the ball take to reach to the height of $30$ m?
OR
(iii) (b) Find the two different values of $t$ when the height of the ball was $20$ m.
Show SolutionHide Solution↓
(i) Zeroes of the polynomial are $0$ and $5$
(ii) Maximum height achieved by ball
$= 25 \times \frac{5}{2} - 5 \times (\frac{5}{2})^2$
$= \frac{125}{4}$ or $31.25$ m
(iii) (a) $-5t^2 + 25t = 30$
$\Rightarrow t^2 - 5t + 6 = 0$
$\Rightarrow (t-2)(t-3) = 0$
$t \ne 3$, $t = 2$
OR
(iii) (b) $-5t^2 + 25t = 20$
$\Rightarrow t^2 - 5t + 4 = 0$
$\Rightarrow (t - 4)(t - 1) = 0$
$\Rightarrow t = 4, 1$
(ii) Maximum height achieved by ball
$= 25 \times \frac{5}{2} - 5 \times (\frac{5}{2})^2$
$= \frac{125}{4}$ or $31.25$ m
(iii) (a) $-5t^2 + 25t = 30$
$\Rightarrow t^2 - 5t + 6 = 0$
$\Rightarrow (t-2)(t-3) = 0$
$t \ne 3$, $t = 2$
OR
(iii) (b) $-5t^2 + 25t = 20$
$\Rightarrow t^2 - 5t + 4 = 0$
$\Rightarrow (t - 4)(t - 1) = 0$
$\Rightarrow t = 4, 1$