Question

2) A particle is projected at an
angle 60 degrees with the horizontal with a
speed lomls. Then the maximum height reached is... knowing that g=10 m/s^2​

Answer 1

Given :   v=20m/s

Given :   v=20m/s∴   ux=vcos60o=20×0.5=10m/s    (which remains the constant all the time)

Given :   v=20m/s∴   ux=vcos60o=20×0.5=10m/s    (which remains the constant all the time)Also   uy=vsin60o=20×0.866=17.32m/s     (which decreases with time)

Given :   v=20m/s∴   ux=vcos60o=20×0.5=10m/s    (which remains the constant all the time)Also   uy=vsin60o=20×0.866=17.32m/s     (which decreases with time)Let the time be t when the final velocity is half of initial velocity  i.e   Vf=220=10m/s

Given :   v=20m/s∴   ux=vcos60o=20×0.5=10m/s    (which remains the constant all the time)Also   uy=vsin60o=20×0.866=17.32m/s     (which decreases with time)Let the time be t when the final velocity is half of initial velocity  i.e   Vf=220=10m/sThus at time t, the y component of the velocity is zero  i.e  Vy=0

Given :   v=20m/s∴   ux=vcos60o=20×0.5=10m/s    (which remains the constant all the time)Also   uy=vsin60o=20×0.866=17.32m/s     (which decreases with time)Let the time be t when the final velocity is half of initial velocity  i.e   Vf=220=10m/sThus at time t, the y component of the velocity is zero  i.e  Vy=0∴   0=uy−gt

Given :   v=20m/s∴   ux=vcos60o=20×0.5=10m/s    (which remains the constant all the time)Also   uy=vsin60o=20×0.866=17.32m/s     (which decreases with time)Let the time be t when the final velocity is half of initial velocity  i.e   Vf=220=10m/sThus at time t, the y component of the velocity is zero  i.e  Vy=0∴   0=uy−gtOR       t=guy=1017.32=1.732  s

hope it helps ..