A body projected with a speed v has its speed v/2 at the highest position of its trajectory. The maximum height attained by it is ?

1 Answer
Jul 7, 2017

h = (3v^2)/(8g)

Explanation:

We know that it was launched with a speed v, and at its highest point, the y-velocity is zero, so the speed of v/2 is solely the x-component (which does not change).

The initial x-velocity is thus also v/2, and the magnitude of the initial velocity is v, so the initial y-velocity is given by the Pythagorean theorem:

v_(0y) = sqrt((v)^2 - (v/2)^2) = (vsqrt3)/2

(which are components of a 30^"o"-60^"o"-90^"o" triangle)

The maximum height h attained is given by the kinematics equation

(v_y)^2 = (v_(0y))^2 - 2gh

Plugging in known values, we have

0 = 0.75v^2 - 2(9.81color(white)(l)"m/s"^2)h

Rearranging and simplifying gives

h = (0.75v^2)/(19.62color(white)(l)"m/s"^2)

We can also simply write it with the acceleration g to make things even simpler:

color(blue)(h = (0.75v^2)/(2g)

or

color(blue)(h = (3v^2)/(8g)