In the most basic of ways. Break it down to the bare minimum. PLEASE!!!
-
sure.
projectile when launched has a velocity v which is inclined at an angle (theta) to the plain
now the vertical component of the velocity(which is v(sin(theta))) is affected by gravity whereas the horizontal component of gravity (v(cos(theta))) remains unaltered.
first we find the time of flight. this would be equal to the time it takes for the vertical component of velocity to get reversed.(this is because when you throw something up with velocity v, it comes down with velocity v). apply v=u+gt
now we will find range. for this simply multiply the horizontal component of velocity with the time of flight.
for calculating height acheived, we apply s=ut+1/2gt^2
projectile when launched has a velocity v which is inclined at an angle (theta) to the plain
now the vertical component of the velocity(which is v(sin(theta))) is affected by gravity whereas the horizontal component of gravity (v(cos(theta))) remains unaltered.
first we find the time of flight. this would be equal to the time it takes for the vertical component of velocity to get reversed.(this is because when you throw something up with velocity v, it comes down with velocity v). apply v=u+gt
now we will find range. for this simply multiply the horizontal component of velocity with the time of flight.
for calculating height acheived, we apply s=ut+1/2gt^2
-
Ask a vague question, get a default answer:
http://en.wikipedia.org/wiki/Projectile_…
http://en.wikipedia.org/wiki/Projectile_…