The equations of motion of two stones thrown | vedclass

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

Question

(D) For the first stone,$s_1 = 19.6t - 4.9t^2$. The velocity is $v_1 = \frac{ds_1}{dt} = 19.6 - 9.8t$. At maximum height,$v_1 = 0$,so $19.6 - 9.8t = 0

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) For the first stone,$s_1 = 19.6t - 4.9t^2$. The velocity is $v_1 = \frac{ds_1}{dt} = 19.6 - 9.8t$. At maximum height,$v_1 = 0$,so $19.6 - 9.8t = 0$,which gives $t = 2 \, s$. The maximum height $h$ is $s_1(2) = 19.6(2) - 4.9(2^2) = 39.2 - 19.6 = 19.6 \, m$. Now,for the second stone,$s_2 = 9.8t - 4.9t^2$. At $t = 2 \, s$,the height of the second stone is $s_2(2) = 9.8(2) - 4.9(2^2) = 19.6 - 19.6 = 0 \, m$. Thus,when the first stone is at its maximum height,the second stone is on the ground.

The equations of motion of two stones thrown vertically upwards simultaneously are $s_1 = 19.6t - 4.9t^2$ and $s_2 = 9.8t - 4.9t^2$ respectively. If the maximum height attained by the first stone is $h$,then when the first stone is at its maximum height,the height of the second stone will be:

Similar Questions

Twenty metres of wire is available for fencing off a flowerbed in the form of a circular sector. Then the maximum area (in sq. m) of the flower bed is:

If $f(x) = 2x^3 - 3x^2 - 12x + 5$ and $x \in [-2, 4]$,then the maximum value of the function occurs at which value of $x$?

Two sides of a triangle are given. If the area of the triangle is maximum,then the angle between the given sides is (in $^{\circ}$)

If $f(x) = 1 + 2 \sin x + 3 \cos^2 x$ for $0 < x < 2\pi / 3$,then:

The number of points in $(-\infty, \infty)$ for which $x^2 - x \sin x - \cos x = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module