Question

2D Kinematics: Projectile Motion

Two balls are thrown up in the air from the same level or floor as shown below in figure. Ball, A, is thrown up vertically in upward direction with velocity (v0). While Ball, B, is thrown up in projectile path with velocity (2v0) at an angle of 300 relative to horizontal axis. Determine which ball flies to the maximum height? Express your calculation in detail.

4

Answer 1

Accoding to the motion in two dimensional

Maximum height on projectile motion H=u^2sin^2(theta)/2g

Given that
initial velocity u1=vo

initial velocity u2=2vo

angle $\theta$ 2=30 degree

angle $\theta$ 1=90 degree

now we find the each maximum height

maximum height for first ball H1=Vo^2sin^2(90)/2g

=vo^2(1)^2/2g

=vo^2/2g..................................(1)

maximum height for the second ball H2=(2vo)^2sin ^2(30)/2g

=4vo^2*(1/2)^2/2g

=4vo^2/4*2g

=vo^2/2g............................(2)

from equation 1 and eq2 based both the balls are reach same height