If a ball is thrown vertically upwards and th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the displacement equation: $s = 22t - 11t^{2}$.To find the maximum height, we differentiate $s$ with respect to $t$ to find the velocity $v

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) Given the displacement equation: $s = 22t - 11t^{2}$.To find the maximum height, we differentiate $s$ with respect to $t$ to find the velocity $v = \frac{ds}{dt}$.$\frac{ds}{dt} = 22 - 22t$.At the maximum height, the velocity is zero, so $22 - 22t = 0$, which gives $t = 1 	ext{ s}$.Substituting $t = 1$ into the displacement equation:$s = 22(1) - 11(1)^{2} = 22 - 11 = 11 	ext{ units}$.Since the ball starts from $s=0$ and reaches a maximum height of $11 	ext{ units}$, the total distance travelled is $11 	ext{ units}$.

If a ball is thrown vertically upwards and the height $s$ reached in time $t$ is given by $s = 22t - 11t^{2}$, then the total distance travelled by the ball is (in $\text{ units}$)

Similar Questions

The minimum value of the polynomial $x(x + 1)(x + 2)(x + 3)$ is

If the function $y = a \log x + bx^2 + x$ has extreme values at $x = 1$ and $x = 2$,then $(a, b) = \dots$

Show that the function given by $f(x) = \frac{\log x}{x}$ has a maximum at $x = e$.

If the extreme value of the function $f(x) = \frac{4}{\sin x} + \frac{1}{1 - \sin x}$ in the interval $[0, \frac{\pi}{2}]$ is $m$ and it exists at $x = k$,then $\cos k =$

The necessary condition for a function to have a maximum or minimum value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module