Question

A ball is lunched at to = 0 with a vertical ascending velocity V_0 of magnitude V_0 = 10m/s from point O and at the instant to = 0. The ball describes rectilinear motion on a vertical axis O_z oriented upwards. we neglect air resistance Given g = 10 m/s Study the motion Determine the parametric eq. of motion. Determine with respect to the position of the top S of the trajectory.

Answer 1

i) the ball ascends and decends to the ground since it has only vertical velocity in z direction. So it will be in free fall motion. i.e under the influence of gravity all the time.

(ii) THe equation of motion for the ball is

$\overrightarrow{z} = \overrightarrow{V_{0}}t + (1/2)\overrightarrow{g}t^2$

So z = 10t - (1/2) * 9.8 *t^2

maximum height is reached when final velocity = 0 implies 0 = 10 - 9.8t so t= 1.02s

So maximum height = 10(1.02) - 4.9 (1.02)^2 = 5.1 metres

Given 1.6 metres above the ground

So ball reaches ground means z = -1.6

So -1.6 = 10t -4.9 t^2

t = 2.18 seconds

velocity = v = u -gt = 10 - 9.8 * 2.18 = -11.364 m/s which means bal hits the ground with a downward speed of 11.36 m/s